Samacheer Class 12 Maths - Inverse Trigonometric Functions

(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.

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.)

(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).

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.

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.

(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.

(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].

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.

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.

We know that tan(tan -1 x) = x
(i) $\tan \left(\tan ^{-1} \frac{7 \pi}{4}\right)=\frac{7 \pi}{4}$
(ii) tan(tan -1 (1947))= 1947
(iii) tan(tan -1 (-0.2021)) = -0.2021

(i) cos^{-1}(1/2)=π/3 and sin^{-1}(-1/2)=-π/6, so the angle is π/2. Therefore tan(π/2) is not defined.

(ii) Let A=tan^{-1}(1/2) and B=cos^{-1}(4/5). Then sin A=1/√5, cos A=2/√5, sin B=3/5, and cos B=4/5. Hence sin(A-B)=sin A cos B - cos A sin B = 4/(5√5)-6/(5√5) = -2/(5√5).

(iii) Let C=sin^{-1}(4/5) and D=tan^{-1}(3/4). Then cos C=3/5, sin C=4/5, cos D=4/5, and sin D=3/5. Hence cos(C-D)=cos C cos D + sin C sin D = 12/25+12/25 = 24/25.

(i) Let $\sec^{-1}\!\left(\tfrac{2}{\sqrt3}\right)=\theta$, so $\sec\theta=\tfrac{2}{\sqrt3}$ with $\theta\in[0,\pi]\setminus\{\tfrac\pi2\}$. Since $\sec\tfrac\pi6=\tfrac{2}{\sqrt3}$, $\theta=\tfrac\pi6$. (ii) Let $\cot^{-1}(\sqrt3)=\theta$, $\theta\in(0,\pi)$; $\cot\tfrac\pi6=\sqrt3$, so $\theta=\tfrac\pi6$. (iii) Let $\operatorname{cosec}^{-1}(-\sqrt2)=\theta$, $\theta\in[-\tfrac\pi2,\tfrac\pi2]\setminus\{0\}$; $\operatorname{cosec}\!\left(-\tfrac\pi4\right)=-\sqrt2$, so $\theta=-\tfrac\pi4$.

(i) $\tan^{-1}(\sqrt3)=\dfrac\pi3$ and $\sec^{-1}(-2)=\dfrac{2\pi}{3}$, so the value is $-\dfrac\pi3$.

(ii) $\sin^{-1}(-1)=-\dfrac\pi2$ and $\cos^{-1}\!\left(\dfrac12\right)=\dfrac\pi3$, so the value is $\cot^{-1}(2)-\dfrac\pi6$.

(iii) $\cot^{-1}(1)=\dfrac\pi4$, $\sin^{-1}\!\left(-\dfrac{\sqrt3}{2}\right)=-\dfrac\pi3$, and $\sec^{-1}(-\sqrt2)=\dfrac{3\pi}{4}$. Therefore the value is $\dfrac\pi4-\dfrac\pi3-\dfrac{3\pi}{4}=-\dfrac{5\pi}{6}$.

Let $A=\tan^{-1}x,\ B=\tan^{-1}y,\ C=\tan^{-1}z$, so $A+B+C=\pi$ and $A+B=\pi-C$. Then $\tan(A+B)=\tan(\pi-C)=-\tan C=-z$. But $\tan(A+B)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}=\dfrac{x+y}{1-xy}$. So $\dfrac{x+y}{1-xy}=-z\Rightarrow x+y=-z(1-xy)=-z+xyz\Rightarrow x+y+z=xyz$.

Put $x=\tan\theta$. Then $\dfrac{2x}{1-x^2}=\dfrac{2\tan\theta}{1-\tan^2\theta}=\tan2\theta$, so $\tan^{-1}\dfrac{2x}{1-x^2}=2\theta$. Hence the left side $=\theta+2\theta=3\theta=3\tan^{-1}x$. Also $\tan3\theta=\dfrac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}=\dfrac{3x-x^3}{1-3x^2}$, so $3\theta=\tan^{-1}\dfrac{3x-x^3}{1-3x^2}$ (the condition $|x|<\tfrac1{\sqrt3}$ keeps $1-3x^2>0$). Therefore the identity holds.