If cotθ=87, evaluate (1+cosθ)(1−cosθ)(1+sinθ)(1−sinθ)

Let us draw a right triangle ABC in which ∠B=90o and ∠C=θWe have, cotθ=87=PerpendicularBase=ABBC (given)Let BC=7k and AB=8kTherefore, by Pythagoras TheoremAC2=AB2+BC2=(8k)2+(7k)2=64k2+49k2AC2=113k2 ∴AC=113k∴sinθ=HypotenusePerpendicular=ACAB=113k8k=1138And cosθ=HypotenuseBase=ACBC=113k7=1137(1+cosθ)(1−cosθ)(1+sinθ)(1−sinθ)=1−cos2θ1−sin2θ=1−(1137)21−(1138)2=1−113491−11364=113113−49113113−64=6449Alternate Method:(1+cosθ)(1−cosθ)(1+sinθ)(1−sinθ)=1−cos2θ1−sin2θ=cot2θ=(87)2=6449

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If ={7}{8}, evaluate {(1+ )(1- )}{(1+)(1- )}

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If $cot θ = \frac{7}{8}$ , evaluate $\frac{( 1 + sin θ ) ( 1 - sin θ )}{( 1 + cos θ ) ( 1 - cos θ )}$

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cot θ = \frac{7}{8}

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

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Text solutionVerified

Let us draw a right triangle

A BC

in which

∠ B = 9 0^{o}

and

∠ C = θ

We have,

cot θ = \frac{7}{8} = \frac{B a se}{P er p e n d i c u l a r} = \frac{BC}{A B}

(given)

Let

BC = 7 k

and

A B = 8 k

Therefore, by Pythagoras Theorem

A C^{2} = A B^{2} + B C^{2} = (8 k)^{2} + (7 k)^{2} = 64 k^{2} + 49 k^{2}

A C^{2} = 113 k^{2}

∴ A C = 113 k

∴ sin θ = \frac{P er p e n d i c u l a r}{Hy p o t e n u se} = \frac{A B}{A C} = \frac{8 k}{113 k} = \frac{8}{113}

And

cos θ = \frac{B a se}{Hy p o t e n u se} = \frac{BC}{A C} = \frac{7}{113 k} = \frac{7}{113}

\frac{( 1 + sin θ ) ( 1 - sin θ )}{( 1 + cos θ ) ( 1 - cos θ )} = \frac{1 - sin ^{2} θ}{1 - cos ^{2} θ} = \frac{1 - ( \frac{8}{113} ) ^{2}}{1 - ( \frac{7}{113} ) ^{2}}

= \frac{1 - \frac{64}{113}}{1 - \frac{49}{113}} = \frac{\frac{113 - 64}{113}}{\frac{113 - 49}{113}} = \frac{49}{64}

Alternate Method:

\frac{( 1 + sin θ ) ( 1 - sin θ )}{( 1 + cos θ ) ( 1 - cos θ )} = \frac{1 - sin ^{2} θ}{1 - cos ^{2} θ} = cot^{2} θ = (\frac{7}{8})^{2} = \frac{49}{64}

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Question Text	If $cot θ = \frac{7}{8}$ , evaluate $\frac{( 1 + sin θ ) ( 1 - sin θ )}{( 1 + cos θ ) ( 1 - cos θ )}$
Answer Type	Text solution:1
Upvotes	150

If cotθ=87​, evaluate (1+cosθ)(1−cosθ)(1+sinθ)(1−sinθ)​

Text solutionVerified

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If $cot θ = \frac{7}{8}$ , evaluate $\frac{( 1 + sin θ ) ( 1 - sin θ )}{( 1 + cos θ ) ( 1 - cos θ )}$