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Trigonometric Ratios of Allied Angles

cos90°+θcosec...

Question

$\frac{cos (90 ° + θ) cosec (270 ° + θ) cos (180 ° + θ)}{sec (- θ) sin (270 ° + θ) sin (360 ° - θ)} =$

A

- 1

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B

1

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C

{cot}^{2} θ

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D

tan θ

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Solution

The correct option is A $- 1$
$\frac{cos (90 ° + θ) cosec (270 ° + θ) cos (180 ° + θ)}{sec (- θ) sin (270 ° + θ) sin (360 ° - θ)}$

$= \frac{- sin θ (- sec θ) (- cos θ)}{sec θ (- cos θ) (- sin θ)} = - 1$

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Trigonometric Ratios of Allied Angles

Standard XIII Mathematics

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