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Q) If cot θ =7/8, then find the value of (1+sin θ)(1−sin θ)/(1+cos θ)(1−cos θ).

(Q 22 B – 30/1/3 – CBSE 2026 Question Paper)

Ans: 

Step 1: Let’s first simiplify the given expression:

\frac{(1+\sin \theta)(1 - \sin \theta)}{(1+\cos \theta)(1 - \cos \theta)}

= \frac{(1-\sin^2 \theta)}{(1 - \cos^2 \theta)} ………… (i)

Step 2: From Trigonometric identity, we know that

sin 2 θ + cos 2 θ = 1

∴ 1 – sin 2 θ = cos 2 θ ………… (ii)

and 1 – cos 2 θ = sin 2 θ ………… (iii)

Step 3: Substituting the values from equation (ii) and (iii), we get:

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

= \frac{(\cos^2 \theta)}{(\sin^2 \theta)}

= cot 2 θ

Step 4: ∵ We are given that cot θ = 7/8

∴ cot 2 θ = (cot θ) 2 = (\frac{7}{8})^2 = \frac{49}{64}

Therefore, the value of given expression is \frac{49}{64}.

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