Samacheer Class 12 Maths Volume 1 - Inverse Trigonometric Functions

Your Progress - Chapter 40% complete

EXERCISE 4.1EXERCISE 4.17 questions

Q.1Find all values of x such that (i) −10π ≤ x ≤ 10π and sin x = 0. (ii) −3π ≤ x ≤ 3π and sin x = −1.v

Solution

(i) sin x = 0 ⇒ x = nπ. With −10π ≤ x ≤ 10π gives n = −10, −9, …, 10. (ii) sin x = −1 ⇒ x = 3π/2 + 2nπ. Choose integers n so that −3π ≤ x ≤ 3π; n = −2, −1, 0 give x = −5π/2, −π/2, 3π/2 respectively.

Answer:

(i) x = kπ, k = −10, −9, …, 0, …, 9, 10. (ii) x = 3π/2 + 2kπ; within the interval these are x = −5π/2, −π/2, 3π/2.

Q.2Find the period and amplitude of (i) y = sin 7x (ii) y = − sin(x/3) (iii) y = −4 sin 2x.v

Solution

For y = A sin(bx): amplitude = |A|, period = 2π/|b|. (i) A=1, b=7 ⇒ period 2π/7. (ii) A=−1, b=1/3 ⇒ period 2π/(1/3)=6π. (iii) A=−4, b=2 ⇒ amplitude 4, period 2π/2=π.

Answer:

(i) Amplitude 1, period 2π/7. (ii) Amplitude 1, period 6π. (iii) Amplitude 4, period π.

Q.3Sketch the graph of y = sin(x/3) for 0 ≤ x < 6π. (Describe key points.)v

Solution

Since period = 6π, on [0,6π) one full cycle. Solve x/3 = 0, π/2, π, 3π/2, 2π ⇒ x = 0, 3π/2, 3π, 9π/2, 6π. Values: 0→0, 3π/2→1, 3π→0, 9π/2→−1, 6π→0. Sketch accordingly.

Answer:

One full sine wave from x=0 to x=6π: zeros at x=0, 3π, 6π; maximum y=1 at x=3π/2; minimum y=−1 at x=9π/2.

Q.4Evaluate (i) sin(sin^{-1}(1/2)) (ii) sin^{-1}(sin(π/5)).v

Solution

(i) For y = sin^{-1}(1/2) principal value lies in [−π/2,π/2] and equals π/6, so sin(sin^{-1}(1/2)) = 1/2. (ii) π/5 ∈ [−π/2,π/2], so sin^{-1}(sin(π/5)) = π/5.

Answer:

(i) 1/2. (ii) π/5.

Q.5For what value of x does sin(sin^{-1} x) = −1 ?v

Solution

sin(sin^{-1} x) = x for all x in [−1,1]. So x = −1 gives equality to −1. Also inverse image of −1 is −1.

Answer:

x = −1.

Q.6Find the domain of (i) f(x) = sin^{-1}(x+1/2) (ii) g(x) = sin^{-1}(2x−1/4).v

Solution

Domain of sin^{-1}(u) is −1 ≤ u ≤ 1. (i) −1 ≤ x+1/2 ≤ 1 ⇒ −3/2 ≤ x ≤ 1/2. (ii) −1 ≤ 2x−1/4 ≤ 1 ⇒ add 1/4: −3/4 ≤ 2x ≤ 5/4 ⇒ divide by 2: −3/8 ≤ x ≤ 5/8. (Note: if original coefficient different, adjust accordingly.)

Answer:

(i) −3/2 ≤ x ≤ 1/2. (ii) (1/4 −1)/2 ≤ x ≤ (1/4 +1)/2 i.e. −3/4 ≤ x ≤ 5/4.

Q.7Find the value of sin(sin^{-1}(5π/9) + cos^{-1}(9π/9)?) (reconstructing: Evaluate sin(sin^{-1}(5/9)+cos^{-1}(9/9))).v

Solution

The printed OCR is unclear; cannot reliably reconstruct a single correct expression. Please supply the clear original expression.

Answer:

Ambiguous as OCR; insufficient clear data.

EXERCISE 4.2EXERCISE 4.28 questions

Q.1Find all values of x such that (i) −6π ≤ x ≤ 6π and cos x = 0. (ii) −5π ≤ x ≤ 5π and cos x = −1.v

Solution

(i) cos x = 0 ⇒ x = π/2 + nπ. Choose n so that −6π ≤ x ≤ 6π. (ii) cos x = −1 ⇒ x = π + 2nπ. Choose n so that −5π ≤ x ≤ 5π; this yields x = −5π, −3π, −π, π, 3π, 5π (as applicable inside interval).

Answer:

(i) x = π/2 + kπ with k integer and −6π ≤ x ≤ 6π ⇒ k = −6, −5, …, 5, 6 giving x = π/2 + kπ. (ii) cos x = −1 ⇒ x = π + 2kπ; within interval x = π + 2kπ with k = −3, −2, −1, 0, 1 gives x = −5π, −3π, −π, π, 3π, 5π? Check: π+2(−3)π= −5π included, up to 5π.

Q.2State the reason why cos(cos^{-1}(−1/6)) ≠ −π/6 (i.e., why you cannot replace cos^{-1} by its argument scaled).v

Solution

The identity is cos(cos^{-1} x) = x for x ∈ [−1,1]. One must not multiply by π; the principal value is an angle in [0,π], but cos of that angle equals the original number x.

Answer:

Because cos^{-1}(u) returns an angle whose cosine is u; cos(cos^{-1}(u)) = u, not the numerical product with π. In particular cos(cos^{-1}(−1/6)) = −1/6, not −π/6.

Q.3Is cos^{-1}(cos(1−x)) = 1−x−π true? Justify.v

Solution

cos^{-1}(cos θ) equals θ if θ ∈ [0,π], otherwise it equals the unique angle in [0,π] with same cosine (which may be |θ| reflected). So formula given is not an identity for all x; it holds only when 1−x ∈ [0,π] and matches the RHS.

Answer:

No in general. cos^{-1}(cos θ) = principal value in [0,π]; equality holds only when θ ∈ [0,π].

Q.4Find the principal value of cos^{-1}(1/1) i.e. cos^{-1}(1).v

Solution

cos θ = 1 ⇒ θ = 0 + 2kπ. The principal value of cos^{-1} lies in [0,π], so cos^{-1}(1)=0.

Answer:

Q.5Evaluate (i) cos^{-1}(1/2)+sin^{-1}(1/2) (ii) cos^{-1}(1)+sin^{-1}(0) (iii) cos^{-1}(cos(7π/17)) + sin^{-1}(sin(7π/17)).v

Solution

(i) cos^{-1}(1/2)=π/3, sin^{-1}(1/2)=π/6 ⇒ sum π/2. (ii) cos^{-1}(1)=0, sin^{-1}(0)=0. (iii) For 7π/17 ≈ 0.41π < π/2 so both inverse-trig return 7π/17; sum = 14π/17.

Answer:

(i) π/3 + π/6 = π/2. (ii) cos^{-1}(1)=0, sin^{-1}(0)=0 ⇒ 0. (iii) 7π/17 + (either 7π/17 or its principal representative) = 14π/17 if both principal values are 7π/17 (and 7π/17 ∈ [0,π/2] so sum = 14π/17).

Q.6Find the domain of (i) f(x)=sin^{-1}(1−x) + cos^{-1}(x/2) (ii) g(x)=sin^{-1}(x)+cos^{-1}(x) (standard).v

Solution

(i) Domain is intersection of domains of the two inverse functions: from 1−x in [−1,1] we get x∈[0,2]; from x/2 in [−1,1] we get x∈[−2,2]. Intersection gives [0,2]. (ii) Both defined for x∈[−1,1].

Answer:

(i) Require −1 ≤ 1−x ≤ 1 ⇒ 0 ≤ x ≤ 2; and −1 ≤ x/2 ≤ 1 ⇒ −2 ≤ x ≤ 2. Intersection: 0 ≤ x ≤ 2. (ii) Domain of both: x ∈ [−1,1].

Q.7For what x does the inequality π/3 < cos^{-1}(1−x) < π/2 hold?v

Solution

cos^{-1}(1−x) ∈ (π/3,π/2) ⇒ 1−x ∈ (cos(π/3), cos(π/2)) = (1/2, 0) but order reversed since cos decreases on [0,π]; properly 0 < 1−x < 1/2 ⇒ 1/2 < x < 1.

Answer:

Solve for 1−x: cos θ decreases on [0,π], so inequality corresponds to cos(π/3) > 1−x > cos(π/2) ⇒ 1/2 > 1−x > 0 ⇒ 1/2 > 1−x > 0 ⇒ 1/2 < x < 1.

Q.8Find (i) cos^{-1}(cos(1/4)) + cos^{-1}(cos(1/5)) (ii) cos^{-1}(cos(1/4)) + cos^{-1}(cos(π/4)). (Assume arguments in radians and values are principal.)v

Solution

For small positive angles θ < π, cos^{-1}(cos θ) = θ. (i) 1/4 and 1/5 are in [0,π], so sum = 1/4+1/5 = 9/20. (ii) 1/4 and π/4 in [0,π], so sum = 1/4 + π/4 = (1+π)/4.

Answer:

(i) 1/4 + 1/5 = 9/20 (since both small positive < π). (ii) 1/4 + π/4 = (1+π)/4 (if 1/4 ∈ [0,π], both principal).

EXERCISE 4.3EXERCISE 4.35 questions

Q.1Find the domain of (i) tan^{-1}(x/2) (?) and (ii) 1/2 + tan^{-1}(1/4 − x) (?) (reconstructed: domain of inverse tangent expressions).v

Solution

tan^{-1}(u) is defined for all real u. Hence any composition with real polynomial inside has domain all real numbers.

Answer:

For arctan expressions the domain is all real numbers. So domain = (−∞, ∞) for both.

Q.2Find (i) tan(tan^{-1}(1/5)) and (ii) tan(tan^{-1}(-1) - tan^{-1}(1)).v

(i) tan(tan^{-1}(1/5)) = 1/5.

(ii) tan^{-1}(-1) = −π/4 and tan^{-1}(1) = π/4, so tan^{-1}(-1) − tan^{-1}(1) = −π/2. tan(−π/2) is undefined, therefore tan(tan^{-1}(-1) − tan^{-1}(1)) is undefined.

Q.3Evaluate (i) tan(tan^{-1}(1) + tan^{-1}(7)?) (ii) tan(tan^{-1}(1/19)+tan^{-1}(47)) (?) (iii) tan(tan^{-1}(1/2021) − tan^{-1}(0)). (Reconstructed common types.)v

Solution

(i) Use tan(A+B) with tan A=1, tan B=7 ⇒ (1+7)/(1−7)=8/(−6)=−4/3. (iii) tan^{-1}(0)=0 so tan(tan^{-1}(1/2021)) = 1/2021. (ii) insufficient clear numeric data to produce unique simplified value here.

Answer:

(i) tan(tan^{-1}(1)+tan^{-1}(7)) = (1+7)/(1−1·7) = 8/(−6) = −4/3. (ii) For small unclear numbers use formula tan(A+B) = (tan A + tan B)/(1 − tan A tan B). (iii) tan(tan^{-1}(1/2021) − tan^{-1}(0)) = 1/2021.

Q.4Evaluate (i) tan^{-1}(cos?) unclear and (ii) sin(tan^{-1}?) unclear. (OCR ambiguous.)v

Solution

The OCR text is too garbled to reconstruct the intended arctan/arcsin compositions precisely. Provide the source expressions for exact evaluation.

Answer:

Ambiguous/insufficiently clear OCR; please provide the exact expressions.

Q.5Evaluate cos(tan^{-1}(1)) − sin(tan^{-1}(1/4)) · tan(tan^{-1}(1/4)) ??? (reconstructed: compute combinations of inverse trig).v

Solution

Use right-triangle relations: for u = tan^{-1} t, tan u = t, sin u = t/√(1+t^2), cos u = 1/√(1+t^2). For t=1: cos=1/√2. For t=1/4: sin=1/√17 and tan=1/4, so their product 1/(4√17). Subtract from 1/√2.

Answer:

If expression is cos(tan^{-1}1) − sin(tan^{-1}1/4)·tan(tan^{-1}1/4): cos(tan^{-1}1)=1/√2, tan(tan^{-1}1/4)=1/4, sin(tan^{-1}1/4)=(1/4)/√(1+(1/4)^2)= (1/4)/√(17/16)=1/√17. Then product = (1/√17)*(1/4)=1/(4√17). So value = 1/√2 − 1/(4√17).

EXERCISE 4.4EXERCISE 4.42 questions

Q.1Find the principal value of (i) sec −       1 2 3 (ii) cot − () 1 3 (iii) cosec − − () 1 2 q 5 2 x x 2 -1 Inverse Trigonometric Functions 155v

Solution

The given text is not parseable unambiguously. I need the precise expressions to compute principal values (each principal value depends on argument domain and the principal range of the inverse function). Please clarify each part.

Answer:

The OCR text is ambiguous and I cannot reliably reconstruct the intended three subproblems. Please re-submit a clear scan / plain-text statement of each part (use standard notation like sec^{-1}(…), cot^{-1}(…) , cosec^{-1}(…)).

Q.2Find the value of (i) tan sec − − () − − () 1 1 3 2 (ii) sin -1 (-1) + cos -1 1      + cot -1 (2) (iii) cot -1 (1) + sin -1 −         2 -sec -1 − () 2v

Solution

Cannot solve until the precise mathematical expressions are provided (current OCR makes the arguments and nesting unclear).

Answer:

The OCR is garbled and I cannot reconstruct the exact expressions. Please provide the three expressions in clear plain text.

EvaluationEvaluation1 questions

Q.1(ii) cos. −       1 1 x = sec − 1 x, if x ∈ − ()  \, 1 1. (iii) tan cot cot. − − −       = > −+<      1 0 0 x x x x x if if p (i) If x ∈ − ()  \, 1 1, then 1 1 1 x ∈ − [], and x ¹ 0. Thus, sin −       1 1 x is well defined. Let sin −       1 1 x = θ. Then, by definition θ π π ∈ −       { } 2 2 0, \ and sin θ = 1 x. Thus, cosec θ = x, which in turn gives θ = − cosec 1 x. Now, sin −       1 1 x = θ = cosec -1 x. Thus, sin − −       = 1 1 1 x x cosec, x ∈ − ()  \, 1 1. Similarly, other results are proved. Property-IV (Reflection identities) (i) sin () sin − − − = − 1 1 x x, if x ∈ − [,] 1 1. (ii) tan () tan − − − = − 1 1 x x, if x ∈ . (iii) cosec cosec − − − = − 1 1 () x x, if x ≥ 1 or x ∈ − ()  \, 1 1. (iv) cos () cos − − − = − 1 1 x x π, if x ∈ − [,] 1 1. (v) sec () sec − − − = − 1 1 x x π, if x ≥ 1 or x ∈ − ()  \, 1 1. (vi) cot () cot − − − = − 1 1 x x π, if x ∈ . (i) If x ∈ − [] 1 1,, then − ∈ − [] x 1 1,. Thus, sin () --1 x is well defined Let sin () − − = 1 x θ. Then θ π π ∈ −       2 2 , and sin θ = − x. Now, sin θ = − x gives x = − = − sin sin() θ θ Inverse Trigonometric Functions 157 From x = − sin() θ, we must have sin − = − 1 x θ, which in turn gives θ = − − sin 1 x. Hence, sin () sin − − − = − 1 1 x x. (iv) If x ∈ − [] 1 1,, then − ∈ − [] x 1 1,. Thus, cos () --1 x is well defined Let cos () − − = 1 x θ. Then θ π ∈[] 0, and cos θ = − x. Now, cos cos cos θ θ π θ = − = − = − () x x implies. Thus, π θ θ π − = = − − − cos, cos 1 1 x x which gives. Hence, cos () cos − − − = − 1 1 x x π. Similarly, other results are proved.v

Solution

Cannot proceed: the text contains multiple statements and proofs; indicate which identity you want proved.

Answer:

The quoted passage appears to be explanatory text about properties of inverse trig functions rather than a single solvable question. If you want a proof of any particular listed identity, please specify which one.

EXERCISE 4.5EXERCISE 4.510 questions

Q.1Find the value, if it exists. If not, give the reason for non-existence. (i) sin cos − () 1 π (ii) tan sin − −             1 5 π (iii) sin sin − [] 1 5.v

Solution

Unable to parse the three parts from the provided OCR. Re-submit clear mathematical expressions.

Answer:

The OCR text is unclear. Please supply each expression clearly (e.g. sin(cos^{-1}(1/π)), tan(sin^{-1}(...)), etc.).

Q.2Find the value of the expression in terms of x, with the help of a reference triangle. (i) sin cos − − () () 1 1 x (iii) cos tan − − () () 1 3 1 x (iii) tan sin −+             1 1 x.v

Solution

Cannot proceed until expressions are clarified.

Answer:

OCR ambiguous. Please re-type the three expressions using standard inverse-trig notation (e.g. sin(cos^{-1}(x)), cos(tan^{-1}(...)), tan(sin^{-1}(...))).

Q.3Find the value of (i) sin cos sin − −                         1 1 3 2 (ii) cot sin sin − − +       1 1 3 5 (iii) tan sin cot − − +       1 1 3v

Solution

Cannot parse the nested inverse functions unambiguously; re-submit clean expressions.

Answer:

Text unreadable. Please provide the three expressions in clear notation.

Q.4Prove that (i) tan tan tan − − − += 1 1 1 2 2 (ii) sin cos sin − − − − = 1 1 1 3v

Solution

Cannot give a proof until the precise identity statements are readable.

Answer:

The OCR is incomplete. If the intended identities are standard addition identities for inverse tangents and composition identities, please resubmit the correctly typed identities (e.g. tan^{-1} x + tan^{-1} y = tan^{-1}((x+y)/(1-xy)) etc.).

Q.5Prove the identity: tan^{-1}x + tan^{-1}y + tan^{-1}z = tan^{-1}\left(\dfrac{x+y+z-xyz}{1-xy-yz-zx}\right) (under the usual conditions on principal values).v

Proof (sketch):
Use the addition formula for tangent twice. Let A = tan^{-1}x, B = tan^{-1}y so tan(A+B) = (x+y)/(1−xy) provided 1−xy ≠ 0. Then
tan(A+B+tan^{-1}z) = ( (x+y)/(1−xy) + z ) / ( 1 − z·(x+y)/(1−xy) ) . Simplify numerator and denominator:

numerator = (x+y+z−xyz)/(1−xy),
denominator = (1−xy−yz−zx)/(1−xy).

Divide numerator by denominator to get (x+y+z−xyz)/(1−xy−yz−zx). Hence
tan^{-1}x + tan^{-1}y + tan^{-1}z = tan^{-1}\left(\dfrac{x+y+z-xyz}{1-xy-yz-zx}\right),
with the caveat that the principal-value branch of tan^{-1} and possible additions of π must be handled so the equality holds in the correct range; the algebraic identity for the tangent of the sum is as given.

Q.6If tan tan tan − − − ++= 1 1 1 x y z π, show that x y z xyz++=.v

Solution

Let A=tan^{-1}x, B=tan^{-1}y, C=tan^{-1}z with A+B+C=π. Then tan(A+B) = tan(π - C) = -tan C = -z. But tan(A+B) = (tan A + tan B)/(1 - tan A tan B) = (x+y)/(1-xy). Therefore (x+y)/(1-xy) = -z ⇒ x+y = -z(1-xy) = -z + xyz ⇒ x+y+z = xyz. Hence proved.

Answer:

OCR missing symbols; likely intended: If tan^{-1}x + tan^{-1}y + tan^{-1}z = π, show that x+y+z = xyz. (This is a standard result.)

Q.7Prove that tan tan tan − − − +− = − − 1 1 1 3 x x x x x x, | | x < 1v

Solution

Cannot provide a precise proof without the exact intended identity. If you mean tan^{-1}x + tan^{-1}(1/x) = π/2 for x>0 (and -π/2 for x<0), proof follows from tan addition formula and sign/quadrant considerations.

Answer:

The OCR is unclear. Possibly the identity is tan^{-1}x + tan^{-1}(1/x) = π/2 for x>0 (or = -π/2 for x<0). If the intended identity is tan^{-1}x - tan^{-1}(x/(1-x^2)) = ? please re-submit the exact formula.

Q.8Simplify: tan tan − − − − + 1 1 x y x y x y.v

Solution

Cannot simplify without the exact readable expression.

Answer:

Presumably the problem asks to simplify tan^{-1}x + tan^{-1}y - tan^{-1}((x+y)/(1-xy)) or a similar combination. The OCR is not clear. Please provide the exact expression in plain text.

Q.9Solve: (i) sin sin − − += 1 1 5 12 2 x x π (ii) 2 1 1 1 tan cos cos − − − = − +− − + x a a b b, a b > > 0 0,. (iii) 2 2 1 1 tan cos tan − − () = () x x cosec (iv) cot cot, − − −+() = > 1 1 2 12 0 x x x π.v

Solution

Cannot solve until the equations are readable.

Answer:

The OCR is too garbled to extract the four separate equations. Please resend each equation in clear plain-text mathematical notation.

Q.10Find the number of solutions of the equation tan tan tan tan − − − − − () +++() = () 1 1 1 1 1 1 3 x x x x.v

Solution

Cannot determine number of solutions until the exact equation is provided clearly.

Answer:

The equation text is not parseable. Please provide the exact equation (use parentheses and standard inverse-trig notation).

Choose the correctChoose the correct19 questions

Q.1 The value of sin^{-1}(cos x), for 0 ≤ x ≤ π is (1) π − x (2) x − π/2 (3) π/2 − x (4) x − π

Answer: (3)

For 0 ≤ x ≤ π, cos x = sin(π/2 - x) and π/2 - x ∈ [-π/2, π/2], the principal range of sin^{-1}. Hence sin^{-1}(cos x) = sin^{-1}(sin(π/2 - x)) = π/2 - x.

Q.2 If sin^{-1}x + sin^{-1}y = π/2 then cos^{-1}x + cos^{-1}y is equal to (1) π/2 (2) π/3 (3) π/6 (4) π

Answer: (1)

cos^{-1}x = π/2 - sin^{-1}x. Therefore cos^{-1}x + cos^{-1}y = π - (sin^{-1}x + sin^{-1}y) = π - π/2 = π/2.

Q.3 sin cos sec − − − − −+− 1 1 1 1 3 cosec is equal to (1) 2 p (2) p (3) 0 (4) tan − 1 12

Answer: Unclear / cannot reconstruct

The OCR rendering of the expression is not clear enough to determine the intended nested inverse functions and their arguments. Please provide the exact expression in plain text so I can evaluate and choose the correct option.

Q.4 If sin^{-1} x = α has a solution, then which of the following is true about α?

Answer: (1)

The principal range of sin^{-1} is [-π/2, π/2], so any α satisfying sin^{-1}x = α must satisfy α ≤ π/2 (and also α ≥ -π/2). Hence option (1) is correct.

Q.5sin(cos^{-1} x) = √(1-x^2) is valid for which x? (options garbled)v

Solution

cos^{-1}x ∈ [0,π], so sin(cos^{-1}x) ≥ 0; hence sin(cos^{-1}x)=+√(1-x^2). This identity is valid for the domain of cos^{-1}, namely -1 ≤ x ≤ 1.

Answer:

sin(cos^{-1} x)=√(1-x^2) for -1 ≤ x ≤ 1

Q.6 If sin^{-1}x + sin^{-1}y + sin^{-1}z = π, the value of expression x y z 2017 2018 2019 101 101 101 ++−++ is (1) 0 (2) 1 (3) 2 (4) 3

Answer: Cannot determine from OCR

The expression to be evaluated is not legible in the OCR (it appears to mix coefficients 2017,2018,2019 and 101,101,101 with x,y,z). Please provide the exact expression in clear text and I will compute the value using the relation derived from α+β+γ=π ⇒ x+y+z=xyz (where x=sinα etc.).

Q.7 If cot^{-1} x = π/2 for some x ∈ R, the value of tan^{-1} x is (1) −π/10 (2) π/5 (3) π/10 (4) −π

Answer: Cannot reconstruct reliably

If cot^{-1}x = π/2 then x = cot(π/2) = 0 and tan^{-1}x = 0. None of the provided options equals 0, so either the OCR mis-copied the target angle or the options. Please confirm the exact problem statement (angle and options).

Q.8 Original OCR (unclear): "The domain of the function defined by f x x ( ) sin = − − 1 1 is (1) 1 2, [] (2) − [] 1 1, (3) 0 1, [] (4) − [] 1 0". Note: OCR is ambiguous. Please re-supply a clear expression for f(x) (likely one of: sin^{-1}((x-1)/(x+1)), sin^{-1}(x-1), sin^{-1}(1-x^2), etc.).

Answer: Cannot determine — question text is ambiguous. Please provide the exact function f(x).

The OCR is not readable enough to reconstruct the precise function. Provide the exact expression (preferably as LaTeX or a clear photo).

Q.10 Original OCR (ambiguous): "tan^{-1}(?) + tan^{-1}(?) = ? 1 1 1 9". Likely intended: "tan^{-1}1 + tan^{-1}(1/9) = ?". Options appear to be inverse cos/sin/tan forms.

Answer: Cannot give a confirmed option due to ambiguous OCR. If the intended sum is tan^{-1}1 + tan^{-1}(1/9), the exact value is tan^{-1}(5/4).

Using tan^{-1}x + tan^{-1}y = tan^{-1}((x+y)/(1-xy)) when xy<1: with x=1, y=1/9, sum = tan^{-1}((1+1/9)/(1-1/9)) = tan^{-1}((10/9)/(8/9)) = tan^{-1}(5/4). Match this to the provided options and choose the corresponding label.

Q.11 Original OCR (ambiguous): "If the function f x x ( ) sin = − () − 1 2 3, then x belongs to ... options show various intervals involving 1/2, -1/2, etc.". Please supply the exact function (likely of the form f(x)=sin^{-1}(x- (something)) or similar).

Answer: Cannot determine — unclear OCR. Please provide the precise expression for f(x).

Text is too garbled to reconstruct the intended inverse-trig function and hence the domain/range question. Re-submit a clear version.

Q.12 If cot^{-1}2 and cot^{-1}3 are two angles of a triangle, then the third angle is: options (1) π/4 (2) 3π/4 (3) π/6 (4) π. (Interpretation: angles are cot^{-1}2, cot^{-1}3, and unknown; sum = π.)

Answer: (1) π/4

Let angles be A = cot^{-1}2, B = cot^{-1}3, C = π - (A+B). Compute tan A = 1/2, tan B = 1/3. Then tan(A+B) = (tanA+tanB)/(1-tanA tanB) = (1/2+1/3)/(1-1/6) = (5/6)/(5/6) =1, so A+B = π/4. Hence C = π - (A+B) = π - π/4 = 3π/4. But careful: inverse cot range gives A,B in (0,π). For triangle angles sum equals π so third angle = π - (A+B) = π - (cot^{-1}2 + cot^{-1}3) = π - (π/4) = 3π/4. However the typical intended result (depending on branch) is 3π/4. Since options include both π/4 and 3π/4, the correct third angle is 3π/4. (Choose option (2)).

Q.13 Original OCR (ambiguous): "sin tan sin − − ( ) − ( ) = 1 1 π x. Then x is a root of the equation ..." The OCR is too garbled to reconstruct the exact identity.

Answer: Cannot determine — please provide a clear statement of the equation involving inverse trigonometric expressions.

OCR is insufficient to reconstruct the trigonometric equation. Re-submit clear text or an image.

Q.14 Original OCR: "sin^{-1}x_1 + cos^{-1}x_2 + cos^{-1}x_1 + sin^{-1}x_2 = ? options: π/2, π/3, π/4, π." (The OCR string: "sin cos cos sin − − − () +− () = 1 2 1 2 2 1 1 2 x x (1) p 2 (2) p 3 (3) p 4 (4) p"). Likely intended identity: sin^{-1}a + cos^{-1}b + cos^{-1}a + sin^{-1}b = ? For a,b in [-1,1]. If a=b, sum = π. If arbitrary but in domain, sum = π.

Answer: (4) π

Using sin^{-1}x + cos^{-1}x = π/2 for any x in [-1,1]. Then sin^{-1}x1 + cos^{-1}x1 = π/2 and cos^{-1}x2 + sin^{-1}x2 = π/2. Sum = π/2 + π/2 = π. Hence option (4).

Q.15 If cot^{-1}(sin α) + tan^{-1}(sin α) = u, then cos 2u equals? (options: tan^2 α, 0, -1, -tan^2 α) — OCR ambiguous about signs.

Answer: (2) 0

Let A = cot^{-1}(sin α), B = tan^{-1}(sin α). Note cot^{-1}x + tan^{-1}x = π/2 for x>0 (and equals -π/2 for x<0) but principal values give A+B = π/2 when sinα≥0. Thus u = π/2 (mod appropriate branch) so cos2u = cosπ = -1? However principal branches: tan^{-1}x ∈ (-π/2,π/2), cot^{-1}x ∈ (0,π). For x in (0,1], cot^{-1}x + tan^{-1}x = π/2, so u = π/2 ⇒ cos2u = cosπ = -1. For x=0, u=π/2 ⇒ cos2u=-1. The multiple-choice most consistent answer is -1 (option (3)). Given branch subtleties, choose (3) -1.

Q.16 Original OCR (ambiguous): "If x ≤ 1, then tan^{-1}x + sin^{-1}x + tan^{-1}x? equals ..." The OCR: "2 2 1 1 tan sin − − −+ x x x is equal to (1) tan^{-1} x (2) sin^{-1} x (3) 0 (4) π". Clarify the exact expression.

Answer: Cannot determine reliably from OCR. Please provide the exact expression to be evaluated.

Ambiguous input. Re-submit clear expression (for example: tan^{-1}x + tan^{-1}(??) or tan^{-1}(something) + sin^{-1}(something)).

Q.17 The equation tan^{-1}x - cot^{-1}x = tan^{-1}(1/x) ??? OCR: "The equation tan cot tan − − − − =       1 1 1 1 x x has ..." Requesting number of solutions.

Answer: Cannot determine from current OCR. Please provide the exact equation in readable form.

The question text is too garbled to identify the equation precisely. Re-submit clearly.

Q.18 Original OCR: "If sin^{-1}(cot^{-1}(1/2) + cot^{-1}(1/2)) = π/?, then x=?" Actually OCR: "If sin cot − − +      = 1 1 1 2 2 x π, then x is equal to (1) 1/2 (2) 1/5 (3) 2/5 (4) 3" — unreadable. Please supply exact expression.

Answer: Cannot determine — provide clear equation.

OCR too noisy. Re-submit clear text or image.

Q.19 Original OCR: "If sin^{-1}(...) + ... = cosec π/4 then value of x is ..." OCR: "If sin − − += 1 1 4 2 x cosec π, then the value of x is (1) 4 (2) 5 (3) 2 (4) 3" — ambiguous.

Answer: Cannot determine — please provide the exact relation involving inverse trig functions.

The OCRed equation is not decipherable. Provide a clear statement.

Q.20 Original OCR: "sin^{-1}(tan x), |x| < 1 is equal to ... options seem to be expressions involving x and sqrt(1-x^2) etc." The OCR shows: "sin(tan), | | − < 1 1 x x is equal to (1) x x 1 2 -(2) 1 1 2 -x (3) 1 1 2 +x (4) x x 1 2 +" but it's unreadable. Please provide the exact intended expression (likely sin^{-1}(tan(sin^{-1}x)) or sin(tan^{-1}x)).

Answer: Cannot determine — the expression is ambiguous. Common related identity: sin^{-1}(tan(sin^{-1}x)) = sin^{-1}(x/√(1−x^2)) for |x|<1, but principal range issues apply.

Please re-type the exact expression (e.g., sin^{-1}(tan x) or sin(tan^{-1}x) or sin^{-1}(tan(sin^{-1}x))). Then a concise solution can be given.

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