NCERT Solutions
Class 10 - Mathematics - Chapter 8: INTRODUCTION TO TRIGONOMETRY - Exercise 8.3

Question 4

Note: In identity proofs, we usually start from one side (LHS) and simplify step-by-step using standard identities until we get the other side (RHS). Since the angles are acute, all values like \(\sin A, \cos A\) are positive and denominators stay non-zero (where defined).

Useful identities (we will use repeatedly):

\(\text{cosec }\theta = \dfrac{1}{\sin\theta}\), \(\sec\theta = \dfrac{1}{\cos\theta}\), \(\cot\theta = \dfrac{\cos\theta}{\sin\theta}\), \(\tan\theta = \dfrac{\sin\theta}{\cos\theta}\)

\(\sin^2\theta + \cos^2\theta = 1\)

\(1+\tan^2\theta = \sec^2\theta\), \(1+\cot^2\theta = \text{cosec }^2\theta\)

(i) Prove: \((\text{cosec } \theta - \cot \theta)^2 = \dfrac{1 - \cos \theta}{1 + \cos \theta}\)

LHS \(= (\text{cosec }\theta-\cot\theta)^2\)

\(= \left(\dfrac{1}{\sin\theta}-\dfrac{\cos\theta}{\sin\theta}\right)^2\)

\(= \left(\dfrac{1-\cos\theta}{\sin\theta}\right)^2\)

\(= \dfrac{(1-\cos\theta)^2}{\sin^2\theta}\)

Now use \(\sin^2\theta = 1-\cos^2\theta = (1-\cos\theta)(1+\cos\theta)\).

So,

\(\dfrac{(1-\cos\theta)^2}{(1-\cos\theta)(1+\cos\theta)} = \dfrac{1-\cos\theta}{1+\cos\theta}\)

\(= \textbf{RHS}\). Hence proved.

(ii) Prove: \(\dfrac{\cos A}{1 + \sin A} + \dfrac{1 + \sin A}{\cos A} = 2 \sec A\)

LHS \(= \dfrac{\cos A}{1+\sin A} + \dfrac{1+\sin A}{\cos A}\)

Take LCM \(= \cos A(1+\sin A)\):

\(= \dfrac{\cos^2 A + (1+\sin A)^2}{\cos A(1+\sin A)}\)

Expand numerator:

\(\cos^2 A + (1 + 2\sin A + \sin^2 A)\)

Group \(\sin^2 A + \cos^2 A = 1\):

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

So,

\(\text{LHS} = \dfrac{2(1+\sin A)}{\cos A(1+\sin A)} = \dfrac{2}{\cos A} = 2\sec A\)

\(= \textbf{RHS}\). Hence proved.

(iii) Prove: \(\dfrac{\tan \theta}{1 - \cot \theta} + \dfrac{\cot \theta}{1 - \tan \theta} = 1 + \sec \theta \text{cosec } \theta\)

LHS \(= \dfrac{\tan\theta}{1-\cot\theta} + \dfrac{\cot\theta}{1-\tan\theta}\)

Write \(\tan\theta=\dfrac{\sin\theta}{\cos\theta}\), \(\cot\theta=\dfrac{\cos\theta}{\sin\theta}\):

First term:

\(\dfrac{\sin\theta/\cos\theta}{1-\cos\theta/\sin\theta} = \dfrac{\sin\theta}{\cos\theta}\cdot\dfrac{\sin\theta}{\sin\theta-\cos\theta} = \dfrac{\sin^2\theta}{\cos\theta(\sin\theta-\cos\theta)}\)

Second term:

\(\dfrac{\cos\theta/\sin\theta}{1-\sin\theta/\cos\theta} = \dfrac{\cos\theta}{\sin\theta}\cdot\dfrac{\cos\theta}{\cos\theta-\sin\theta} = \dfrac{\cos^2\theta}{\sin\theta(\cos\theta-\sin\theta)}\)

Note \((\cos\theta-\sin\theta)=-(\sin\theta-\cos\theta)\), so:

Second term \(= -\dfrac{\cos^2\theta}{\sin\theta(\sin\theta-\cos\theta)}\)

Now add:

\(\text{LHS}=\dfrac{\sin^2\theta}{\cos\theta(\sin\theta-\cos\theta)} - \dfrac{\cos^2\theta}{\sin\theta(\sin\theta-\cos\theta)}\)

Take common denominator \(\sin\theta\cos\theta(\sin\theta-\cos\theta)\):

\(= \dfrac{\sin^3\theta - \cos^3\theta}{\sin\theta\cos\theta(\sin\theta-\cos\theta)}\)

Use \(a^3-b^3=(a-b)(a^2+ab+b^2)\) with \(a=\sin\theta\), \(b=\cos\theta\):

\(\sin^3\theta-\cos^3\theta=(\sin\theta-\cos\theta)(\sin^2\theta+\sin\theta\cos\theta+\cos^2\theta)\)

Cancel \((\sin\theta-\cos\theta)\):

\(\text{LHS}=\dfrac{\sin^2\theta+\sin\theta\cos\theta+\cos^2\theta}{\sin\theta\cos\theta}\)

Replace \(\sin^2\theta+\cos^2\theta=1\):

\(= \dfrac{1+\sin\theta\cos\theta}{\sin\theta\cos\theta} = \dfrac{1}{\sin\theta\cos\theta}+1\)

But \(\dfrac{1}{\sin\theta\cos\theta}=\dfrac{1}{\sin\theta}\cdot\dfrac{1}{\cos\theta}=\text{cosec }\theta\sec\theta\).

So, \(\text{LHS}=1+\sec\theta\text{cosec }\theta=\textbf{RHS}\). Hence proved.

(iv) Prove: \(\dfrac{1 + \sec A}{\sec A} = \dfrac{\sin^2 A}{1 - \cos A}\)

LHS \(= \dfrac{1+\sec A}{\sec A} = \dfrac{1}{\sec A} + \dfrac{\sec A}{\sec A}\)

\(= \cos A + 1 = 1+\cos A\)

RHS \(= \dfrac{\sin^2 A}{1-\cos A}\)

Use \(\sin^2 A = 1-\cos^2 A = (1-\cos A)(1+\cos A)\):

\(\text{RHS}=\dfrac{(1-\cos A)(1+\cos A)}{1-\cos A}=1+\cos A\)

So LHS = RHS. Hence proved.

(v) Prove: \(\dfrac{\cos A - \sin A + 1}{\cos A + \sin A - 1} = \text{cosec } A + \cot A\)

LHS \(= \dfrac{1+\cos A-\sin A}{\cos A+\sin A-1}\)

Multiply numerator and denominator by \((1+\cos A+\sin A)\) (a helpful conjugate-type factor):

\(\text{LHS}=\dfrac{(1+\cos A-\sin A)(1+\cos A+\sin A)}{(\cos A+\sin A-1)(1+\cos A+\sin A)}\)

Numerator is of form \((X-Y)(X+Y)=X^2-Y^2\) with \(X=1+\cos A\), \(Y=\sin A\):

Numerator \(=(1+\cos A)^2-\sin^2 A\)

\(=1+2\cos A+\cos^2 A-\sin^2 A\)

\(=1+2\cos A+(\cos^2 A-\sin^2 A)\)

But \(\cos^2 A-\sin^2 A = (\cos^2 A+\sin^2 A)-2\sin^2 A = 1-2\sin^2 A\). So numerator:

\(=1+2\cos A+1-2\sin^2 A = 2+2\cos A-2\sin^2 A\)

\(=2(1+\cos A-\sin^2 A)\)

Now use \(\sin^2 A = 1-\cos^2 A\):

\(1+\cos A-\sin^2 A = 1+\cos A-(1-\cos^2 A)=\cos A+\cos^2 A=\cos A(1+\cos A)\)

So numerator \(=2\cos A(1+\cos A)\).

Denominator:

\((\cos A+\sin A-1)(1+\cos A+\sin A)=(\cos A+\sin A)^2-1\)

\(=\cos^2 A+\sin^2 A+2\sin A\cos A-1\)

\(=1+2\sin A\cos A-1=2\sin A\cos A\)

Therefore,

\(\text{LHS}=\dfrac{2\cos A(1+\cos A)}{2\sin A\cos A}=\dfrac{1+\cos A}{\sin A}=\dfrac{1}{\sin A}+\dfrac{\cos A}{\sin A}\)

\(=\text{cosec } A+\cot A=\textbf{RHS}\). Hence proved.

(vi) Prove: \(\sqrt{\dfrac{1 + \sin A}{1 - \sin A}} = \sec A + \tan A\)

Start with LHS:

\(\sqrt{\dfrac{1+\sin A}{1-\sin A}}\)

Multiply inside by \(\dfrac{1+\sin A}{1+\sin A}\):

\(=\sqrt{\dfrac{(1+\sin A)^2}{1-\sin^2 A}}\)

But \(1-\sin^2 A = \cos^2 A\). So:

\(=\sqrt{\dfrac{(1+\sin A)^2}{\cos^2 A}}\)

For acute angles, \(\cos A>0\), so:

\(=\dfrac{1+\sin A}{\cos A}=\dfrac{1}{\cos A}+\dfrac{\sin A}{\cos A}=\sec A+\tan A\)

\(=\textbf{RHS}\). Hence proved.

(vii) Prove: \(\dfrac{\sin \theta - 2 \sin^3 \theta}{2 \cos^3 \theta - \cos \theta} = \tan \theta\)

LHS \(=\dfrac{\sin\theta-2\sin^3\theta}{2\cos^3\theta-\cos\theta}\)

Factor numerator and denominator:

Numerator \(=\sin\theta(1-2\sin^2\theta)\)

Denominator \(=\cos\theta(2\cos^2\theta-1)\)

So,

\(\text{LHS}=\dfrac{\sin\theta}{\cos\theta}\cdot\dfrac{1-2\sin^2\theta}{2\cos^2\theta-1}\)

Now use \(\sin^2\theta=1-\cos^2\theta\):

\(1-2\sin^2\theta=1-2(1-\cos^2\theta)=1-2+2\cos^2\theta=2\cos^2\theta-1\)

So the fraction becomes 1, hence:

\(\text{LHS}=\dfrac{\sin\theta}{\cos\theta}=\tan\theta=\textbf{RHS}\). Hence proved.

(viii) Prove: \((\sin A + \text{cosec } A)^2 + (\cos A + \sec A)^2 = 7 + \tan^2 A + \cot^2 A\)

LHS \(= (\sin A + \text{cosec } A)^2 + (\cos A + \sec A)^2\)

Expanding both squares:

\((\sin A + \text{cosec } A)^2 = \sin^2 A + \text{cosec }^2 A + 2\sin A\text{cosec } A\)

Since \(\sin A\text{cosec } A = 1\), this becomes:

\((\sin A + \text{cosec } A)^2 = \sin^2 A + \text{cosec }^2 A + 2\)

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

Since \(\cos A\sec A = 1\), we get:

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

Adding both expressions:

\(\text{LHS} = (\sin^2 A + \text{cosec }^2 A + 2) + (\cos^2 A + \sec^2 A + 2)\)

\(= (\sin^2 A + \cos^2 A) + (\sec^2 A + \text{cosec }^2 A) + 4\)

Using identities \(\sin^2 A + \cos^2 A = 1\), \(\sec^2 A = 1 + \tan^2 A\), and \(\text{cosec }^2 A = 1 + \cot^2 A\):

\(\text{LHS} = 1 + (1 + \tan^2 A) + (1 + \cot^2 A) + 4\)

\(= 7 + \tan^2 A + \cot^2 A\)

Hence proved:

\((\sin A + \text{cosec } A)^2 + (\cos A + \sec A)^2 = 7 + \tan^2 A + \cot^2 A\)

(ix) Prove: \((\text{cosec } A - \sin A)(\sec A - \cos A) = \dfrac{1}{\tan A + \cot A}\)

LHS \(=(\text{cosec } A-\sin A)(\sec A-\cos A)\)

Write in \(\sin,\cos\):

\(=\left(\dfrac{1}{\sin A}-\sin A\right)\left(\dfrac{1}{\cos A}-\cos A\right)\)

\(=\left(\dfrac{1-\sin^2 A}{\sin A}\right)\left(\dfrac{1-\cos^2 A}{\cos A}\right)\)

Use \(1-\sin^2 A=\cos^2 A\) and \(1-\cos^2 A=\sin^2 A\):

\(=\left(\dfrac{\cos^2 A}{\sin A}\right)\left(\dfrac{\sin^2 A}{\cos A}\right)=\sin A\cos A\)

RHS \(=\dfrac{1}{\tan A+\cot A}=\dfrac{1}{\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}}\)

\(=\dfrac{1}{\frac{\sin^2 A+\cos^2 A}{\sin A\cos A}}=\dfrac{1}{\frac{1}{\sin A\cos A}}=\sin A\cos A\)

So LHS = RHS. Hence proved.

(x) Prove: \(\dfrac{1 + \tan^2 A}{1 + \cot^2 A} = \left( \dfrac{1 - \tan A}{1 - \cot A} \right)^2 = \tan^2 A\)

Part 1: \(\dfrac{1+\tan^2 A}{1+\cot^2 A}\)

Use \(1+\tan^2 A=\sec^2 A\) and \(1+\cot^2 A=\text{cosec }^2 A\):

\(=\dfrac{\sec^2 A}{\text{cosec }^2 A}=\dfrac{\frac{1}{\cos^2 A}}{\frac{1}{\sin^2 A}}=\dfrac{\sin^2 A}{\cos^2 A}=\tan^2 A\)

Part 2: \(\left(\dfrac{1-\tan A}{1-\cot A}\right)^2\)

Inside the bracket, write \(\cot A=\dfrac{1}{\tan A}\). Let \(t=\tan A\) (\(t\neq 0\)):

\(\dfrac{1-t}{1-\frac{1}{t}}=\dfrac{1-t}{\frac{t-1}{t}}=(1-t)\cdot\dfrac{t}{t-1}\)

But \(1-t=-(t-1)\), so:

\(=( -(t-1))\cdot\dfrac{t}{t-1}=-t\)

Now square it:

\(\left(-t\right)^2=t^2=\tan^2 A\)

So both expressions equal \(\tan^2 A\). Hence proved.

Final Note: In proofs, always keep denominators non-zero and use identities like \(\sin^2+\cos^2=1\), \(1+\tan^2=\sec^2\), and \(1+\cot^2=\text{cosec }^2\) to simplify quickly.