If cotθ=87, evaluate cot2θ

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If ={7}{8}, evaluate ^{2}

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If $cot θ = \frac{7}{8}$ , evaluate $cot^{2} θ$

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cot θ = \frac{7}{8}

, evaluate

cot^{2} θ

10 mins ago

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10 mins ago

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}

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

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