NCERT Solutions: Class 10 Maths Chapter 8 - Introduction to Trigonometry

By Pythagoras theorem, $AC=\sqrt{24^2+7^2}=25$. For angle A, opposite side is BC and adjacent side is AB. For angle C, opposite side is AB and adjacent side is BC.

In the right triangle of Fig. 8.13, angles P and R are complementary. Therefore $\tan P=\cot R$, so $\tan P-\cot R=0$.

Take opposite side $=3k$ and hypotenuse $=4k$. Then adjacent side $=\sqrt{(4k)^2-(3k)^2}=\sqrt7k$. Hence $\cos A=\dfrac{\sqrt7}{4}$ and $\tan A=\dfrac{3}{\sqrt7}$.

$15\cot A=8$, so $\cot A=\dfrac{8}{15}$. Take adjacent side $=8k$ and opposite side $=15k$; the hypotenuse is $17k$. Therefore $\sin A=\dfrac{15}{17}$ and $\sec A=\dfrac{17}{8}$.

$\sec\theta=\dfrac{13}{12}$ gives hypotenuse $=13k$ and adjacent side $=12k$. The opposite side is $\sqrt{13^2-12^2}k=5k$. The remaining ratios follow from their definitions.

For acute angles, the value of cosine decreases uniquely as the angle increases from $0^\circ$ to $90^\circ$. Hence two acute angles with the same cosine must be equal. Therefore $\angle A=\angle B$.

$\dfrac{(1+\sin\theta)(1-\sin\theta)}{(1+\cos\theta)(1-\cos\theta)}=\dfrac{1-\sin^2\theta}{1-\cos^2\theta}=\dfrac{\cos^2\theta}{\sin^2\theta}=\cot^2\theta$. Since $\cot\theta=\dfrac78$, both values are $\dfrac{49}{64}$.

$3\cot A=4$ gives $\cot A=\dfrac43$, so $\tan A=\dfrac34$. LHS $=\dfrac{1-\frac{9}{16}}{1+\frac{9}{16}}=\dfrac{7}{25}$. Taking sides $3,4,5$, $\sin A=\dfrac35$ and $\cos A=\dfrac45$, so RHS $=\dfrac{16}{25}-\dfrac{9}{25}=\dfrac{7}{25}$.

$\tan A=\dfrac{1}{\sqrt3}$ gives $A=30^\circ$ and, since B is $90^\circ$, $C=60^\circ$. Thus (i) $\sin A\cos C+\cos A\sin C=\sin(A+C)=\sin90^\circ=1$, and (ii) $\cos A\cos C-\sin A\sin C=\cos(A+C)=\cos90^\circ=0$.

Let $QR=x$, so $PR=25-x$. Since the triangle is right-angled at Q, $(25-x)^2=x^2+5^2$. This gives $x=12$ and $PR=13$. For angle P, opposite side is QR and adjacent side is PQ, so the ratios are $\dfrac{12}{13}$, $\dfrac{5}{13}$ and $\dfrac{12}{5}$.

(i) For example, $\tan60^\circ=\sqrt3\gt1$. (ii) This is possible because $\sec A\ge1$ for acute A; equivalently $\cos A=\dfrac{5}{12}$ is possible. (iii) cos means cosine, not cosecant. (iv) cot A is one trigonometric ratio. (v) $\sin\theta$ cannot exceed 1.

Substitute the standard values: $\sin30^\circ=\cos60^\circ=\dfrac12$, $\sin60^\circ=\cos30^\circ=\dfrac{\sqrt3}{2}$, $\tan45^\circ=1$, $\sec30^\circ=\dfrac{2}{\sqrt3}$ and $\cosec60^\circ=\dfrac{2}{\sqrt3}$. Simplifying gives the listed values.

1. A. As listed in the question.
2. B. As listed in the question.
3. C. As listed in the question.
4. D. As listed in the question.

(i) With $\tan30^\circ=\dfrac1{\sqrt3}$, the value is $\dfrac{\sqrt3}{2}=\sin60^\circ$. (ii) Since $\tan45^\circ=1$, the value is 0. (iii) $\sin2A=2\sin A$ holds among the options for $A=0^\circ$. (iv) The value is $\sqrt3=\tan60^\circ$.

$\tan(A+B)=\sqrt3$ gives $A+B=60^\circ$. Also $\tan(A-B)=\dfrac1{\sqrt3}$ gives $A-B=30^\circ$. Solving, $2A=90^\circ$, so $A=45^\circ$ and $B=15^\circ$.

(i) For $A=B=30^\circ$, $\sin60^\circ\ne\sin30^\circ+\sin30^\circ$. (ii) In the acute-angle table, sine increases from 0 to 1. (iii) Cosine decreases from 1 to 0. (iv) Sine equals cosine only at $45^\circ$ in this range, not for all values. (v) $\cot0^\circ=\dfrac{\cos0^\circ}{\sin0^\circ}$ is not defined because $\sin0^\circ=0$.

Using $\cosec^2 A=1+\cot^2 A$, we get $\cosec A=\sqrt{1+\cot^2 A}$ and hence $\sin A=\dfrac1{\sqrt{1+\cot^2 A}}$. Also $\tan A=\dfrac1{\cot A}$ and $\sec A=\dfrac1{\cos A}=\dfrac{\sqrt{1+\cot^2 A}}{\cot A}$.

From $\sec^2 A=1+\tan^2 A$, $\tan A=\sqrt{\sec^2 A-1}$ for acute A. Also $\cos A=1/\sec A$, $\sin A=\tan A\cos A$, and the reciprocal ratios give cosec A and cot A.

1. A. As listed in the question.
2. B. As listed in the question.
3. C. As listed in the question.
4. D. As listed in the question.

(i) $9(\sec^2A-\tan^2A)=9$. (ii) The product simplifies to 2 using reciprocal identities. (iii) $(\sec A+\tan A)(1-\sin A)=\dfrac{1+\sin A}{\cos A}(1-\sin A)=\cos A$. (iv) $\dfrac{1+\tan^2A}{1+\cot^2A}=\dfrac{\sec^2A}{\cosec^2A}=\tan^2A$.

(i) LHS $=\left(\dfrac{1-\cos\theta}{\sin\theta}\right)^2=\dfrac{(1-\cos\theta)^2}{1-\cos^2\theta}=\dfrac{1-\cos\theta}{1+\cos\theta}$.
(ii) LHS $=\dfrac{\cos^2A+(1+\sin A)^2}{\cos A(1+\sin A)}=\dfrac{2(1+\sin A)}{\cos A(1+\sin A)}=2\sec A$.
(iii) Write in sin and cos; the LHS becomes $\dfrac{\sin^2\theta}{\cos\theta(\sin\theta-\cos\theta)}-\dfrac{\cos^2\theta}{\sin\theta(\sin\theta-\cos\theta)}=1+\sec\theta\cosec\theta$.
(iv) LHS $=1+\cos A$ and RHS $=\dfrac{1-\cos^2A}{1-\cos A}=1+\cos A$.
(v) Simplifying the LHS gives $\dfrac{1+\cos A}{\sin A}=\cosec A+\cot A$.
(vi) $\sqrt{\dfrac{1+\sin A}{1-\sin A}}=\sqrt{\dfrac{(1+\sin A)^2}{\cos^2A}}=\dfrac{1+\sin A}{\cos A}=\sec A+\tan A$.
(vii) Factor numerator and denominator: $\dfrac{\sin\theta(1-2\sin^2\theta)}{\cos\theta(2\cos^2\theta-1)}=\dfrac{\sin\theta(2\cos^2\theta-1)}{\cos\theta(2\cos^2\theta-1)}=\tan\theta$.
(viii) Expanding gives $\sin^2A+\cos^2A+\cosec^2A+\sec^2A+4=1+(1+\cot^2A)+(1+\tan^2A)+4=7+\tan^2A+\cot^2A$.
(ix) LHS $=\left(\dfrac{1-\sin^2A}{\sin A}\right)\left(\dfrac{1-\cos^2A}{\cos A}\right)=\sin A\cos A$, and RHS $=\dfrac{1}{\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}}=\sin A\cos A$.
(x) $\dfrac{1+\tan^2A}{1+\cot^2A}=\dfrac{\sec^2A}{\cosec^2A}=\tan^2A$, and $\dfrac{1-\tan A}{1-\cot A}=\dfrac{1-t}{1-1/t}=-t$ for $t=\tan A$, so its square is $\tan^2A$.