If theta is an acute angle such that tan^2theta=frac{8}{7}, then the value of frac{(1+sintheta)(1-sintheta)}{(1+costheta)(1-costheta)} is:

Q: If $\theta$ is an acute angle such that $\tan^2\theta=\frac{8}{7},$ then the value of $\frac{(1+\sin\theta)(1-\sin\theta)}{(1+\cos\theta)(1-\cos\theta)}$ is:

Given that: $\tan^2\theta=\frac{8}{7}$ and $\theta$ is an acute angle We have to find the following expression $\frac{(1+\sin\theta)(1-\sin\theta)}{(1+\cos\theta)(1-\cos\theta)}$ Since $\tan^2\theta=\frac{8}{7}$ $\tan^2\theta=\sqrt\frac{8}{7}$ $\tan\theta=\frac{\sqrt8}{\sqrt7}$ Since $\tan\theta=\frac{\text{Perpedicular}}{\text{Base}}$ $\Rightarrow{\text{perpedicular}}=\sqrt{8}$ $\Rightarrow\text{Base}=\sqrt{7}$ $\Rightarrow\text{Hypotenuse}=\sqrt{8+7}$ $\Rightarrow\text{Hypotenuse}=\sqrt{15}$ We know that $\sin\theta=\frac{\text{Perpendicular}}{\text{Hypotenuse}}$ and $\cos\theta=\frac{\text{Base}}{\text{Hypotenuse}}$ We find: $\frac{(1+\sin\theta)(1-\sin\theta)}{(1+\cos\theta)(1-\cos\theta)}$ $=\frac{\Big(1+\frac{\sqrt{8}}{\sqrt{15}}\Big)\Big(1-\frac{\sqrt{8}}{\sqrt{15}}\Big)}{\Big(1+\frac{\sqrt{7}}{\sqrt{15}}\Big)\Big(1-\frac{\sqrt{7}}{\sqrt{15}}\Big)}$ $=\frac{\Big(1-\frac{8}{15}\Big)}{\Big(1-\frac{7}{15}\Big)}$ $=\frac{\frac{7}{15}}{\frac{8}{15}}$ $=\frac{7}{8}$ Hence the correct option is $(a)$.

M.C.Q (1 Marks)

GSEB - English Medium

If $\theta$ is an acute angle such that $\tan^2\theta=\frac{8}{7},$ then the value of $\frac{(1+\sin\theta)(1-\sin\theta)}{(1+\cos\theta)(1-\cos\theta)}$ is:

A$\frac{7}{8}$
B$\frac{8}{7}$
C$\frac{7}{4}$
D$\frac{64}{49}$

STD 10
Maths
Trigonometric Ratios
R.D. Sharma

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