Class 10 Mathematics – Chapter 8: Introduction to Trigonometry

Exercise 8.3 – NCERT Solutions

Q1. Express sin A, sec A and tan A in terms of cot A.

Using \(1+\cot^2A=\cosec^2A\),

\(\cosec A=\sqrt{1+\cot^2A}\).

Hence, \(\sin A=\frac1{\sqrt{1+\cot^2A}}\).

Also, \(\tan A=\frac1{\cot A}\).

Therefore, \(\sec A=\sqrt{1+\tan^2A}=\frac{\sqrt{1+\cot^2A}}{\cot A}\), since A is acute.

Q2. Write all the other trigonometric ratios of ∠A in terms of sec A.

Let \(\sec A=s\). Then \(\cos A=\frac1s\).

Using \(\sin^2A+\cos^2A=1\),

\(\sin A=\frac{\sqrt{s^2-1}}s\).

Therefore,

\(\tan A=\sqrt{s^2-1},\quad \cot A=\frac1{\sqrt{s^2-1}},\)

\(\cosec A=\frac{s}{\sqrt{s^2-1}}\).

Q3. Choose the correct option. Justify your choice.

(i) \(9\sec^2A-9\tan^2A=9(\sec^2A-\tan^2A)=9\). Option (B).

(ii) Let \(x=\sec\theta+\tan\theta\). Since \((\sec\theta+\tan\theta)(\sec\theta-\tan\theta)=1\), the expression simplifies to 2. Option (C).

(iii) \((\sec A+\tan A)(1-\sin A)=\frac{1+\sin A}{\cos A}(1-\sin A)=\cos A\). Option (D).

(iv) \(\frac{1+\tan^2A}{1+\cot^2A}=\frac{\sec^2A}{\cosec^2A}=\tan^2A\). Option (D).

Q4(i). Prove \((\cosec\theta-\cot\theta)^2=\frac{1-\cos\theta}{1+\cos\theta}\).

\(\text{LHS}=\left(\frac1{\sin\theta}-\frac{\cos\theta}{\sin\theta}\right)^2=\left(\frac{1-\cos\theta}{\sin\theta}\right)^2\)

\(=\frac{(1-\cos\theta)^2}{(1-\cos\theta)(1+\cos\theta)}=\frac{1-\cos\theta}{1+\cos\theta}=\text{RHS}\).

Q4(ii). Prove \(\frac{\cos A}{1+\sin A}+\frac{1+\sin A}{\cos A}=2\sec A\).

Taking LCM,

\(\text{LHS}=\frac{\cos^2A+(1+\sin A)^2}{\cos A(1+\sin A)}\).

\(=\frac{1-\sin^2A+1+2\sin A+\sin^2A}{\cos A(1+\sin A)}=\frac{2(1+\sin A)}{\cos A(1+\sin A)}=2\sec A\).

Q4(iii). Prove \(\frac{\tan\theta}{1-\cot\theta}+\frac{\cot\theta}{1-\tan\theta}=1+\sec\theta\cosec\theta\).

Put \(t=\tan\theta\). Then \(\cot\theta=1/t\).

\(\text{LHS}=\frac{t}{1-1/t}+\frac{1/t}{1-t}=\frac{t^2}{t-1}-\frac1{t(t-1)}\)

\(=\frac{t^3-1}{t(t-1)}=t+1+\frac1t\).

Since \(\tan\theta+\cot\theta=\sec\theta\cosec\theta\), LHS = RHS.

Q4(iv). Prove \(\frac{1+\sec A}{\sec A}=\frac{\sin^2A}{1-\cos A}\).

\(\text{LHS}=1+\cos A\).

Also, \(\text{RHS}=\frac{1-\cos^2A}{1-\cos A}=1+\cos A\).

Hence LHS = RHS.

Q4(v). Prove \(\frac{\cos A-\sin A+1}{\cos A+\sin A-1}=\cosec A+\cot A\).

Multiply numerator and denominator on the left by \(\cos A-\sin A+1\). Using \(\sin^2A+\cos^2A=1\), the expression simplifies to \(\frac{1+\cos A}{\sin A}\).

But \(\frac{1+\cos A}{\sin A}=\cosec A+\cot A\). Hence proved.

Q4(vi). Prove \(\sqrt{\frac{1+\sin A}{1-\sin A}}=\sec A+\tan A\).

\(\text{LHS}=\sqrt{\frac{(1+\sin A)^2}{1-\sin^2A}}=\sqrt{\frac{(1+\sin A)^2}{\cos^2A}}\).

Since A is acute, this becomes \(\frac{1+\sin A}{\cos A}=\sec A+\tan A\).

Q4(vii). Prove \(\frac{\sin\theta-2\sin^3\theta}{2\cos^3\theta-\cos\theta}=\tan\theta\).

\(\text{LHS}=\frac{\sin\theta(1-2\sin^2\theta)}{\cos\theta(2\cos^2\theta-1)}\).

Since \(1-2\sin^2\theta=2\cos^2\theta-1\), the common factor cancels.

Thus \(\text{LHS}=\frac{\sin\theta}{\cos\theta}=\tan\theta\).

Q4(viii). Prove \((\sin A+\cosec A)^2+(\cos A+\sec A)^2=7+\tan^2A+\cot^2A\).

Expanding the left side,

\(=\sin^2A+\cosec^2A+2+\cos^2A+\sec^2A+2\)

\(=5+\cosec^2A+\sec^2A\).

Using \(\cosec^2A=1+\cot^2A\) and \(\sec^2A=1+\tan^2A\),

\(=7+\tan^2A+\cot^2A\). Hence proved.

Q4(ix). Prove \((\cosec A-\sin A)(\sec A-\cos A)=\frac1{\tan A+\cot A}\).

\(\text{LHS}=\left(\frac{1-\sin^2A}{\sin A}\right)\left(\frac{1-\cos^2A}{\cos A}\right))

\(=\frac{\cos^2A}{\sin A}\cdot\frac{\sin^2A}{\cos A}=\sin A\cos A\).

Also, \(\frac1{\tan A+\cot A}=\frac1{\frac{\sin^2A+\cos^2A}{\sin A\cos A}}=\sin A\cos A\). Hence proved.

Q4(x). Prove \(\left(\frac{1+\tan^2A}{1+\cot^2A}\right)=\left(\frac{1-\tan A}{1-\cot A}\right)^2=\tan^2A\).

First,

\(\frac{1+\tan^2A}{1+\cot^2A}=\frac{\sec^2A}{\cosec^2A}=\tan^2A\).

Also,

\(\frac{1-\tan A}{1-\cot A}=\frac{1-\tan A}{1-1/\tan A}=-\tan A\).

Squaring gives \(\tan^2A\). Therefore both expressions are equal to \(\tan^2A\).