Question

cot°=7\8 then evaluate (1+sin°) 1-sin°)\(1+cos°)(1- cos°). and 1+sin°\cos°=

pls find the answer frnds
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Answer 1

Question :-

If $cot\theta=\frac{7}{8}$ then evaluate :-

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

(2) $\frac{1+sin\theta}{cos\theta}$

Given,

$cot\theta=\frac{7}{8}$

1st Part :-

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

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

$=>\frac{1-sin^{2}\theta}{1-cos^{2}\theta}$

$=>\frac{cos^{2}\theta}{sin^{2}\theta}$

$=>cot^{2}\theta$

$=>(\frac{7}{8} )^{2}$

$=>\frac{49}{64}$

Hence,

The value of $\frac{(1+sin\theta)(1-sin\theta)}{(1+cos\theta)(1-cos\theta)}$ is $\frac{49}{64}$.

2nd Part :-

$\frac{1+sin\theta}{cos\theta}$

$=>\frac{1}{cos\theta} +\frac{sin\theta}{cos\theta}$

$=>\frac{1}{cos\theta} +tan\theta$

$=>\frac{\sqrt{133} }{7} +\frac{8}{7}$

$=>\frac{\sqrt{133} +8}{7}$

Hence,

The value of $\frac{1+sin\theta}{cos\theta}$ is $\frac{\sqrt{133} +8}{7}$.

☆ Hope it Helps ☆

$@NightRead$

Answer 2

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