Trigonometry! #51

Geometry Level 2

Which of the options is equivalent to the following expression? cot ⁡ θ − sin ⁡ ( π 2 − θ ) cos ⁡ ( π 2 − θ ) \cot \theta - \sin \left (\frac {\pi}{2} -\theta \right) \cos \left (\frac {\pi}{2} - \theta \right)

This problem is part of the set Trigonometry .

tan ⁡ 2 θ \tan ^{2} \theta cot ⁡ 2 θ \cot ^{2} \theta sin ⁡ θ cos ⁡ 2 θ \sin \theta \cos ^{2} \theta cot ⁡ θ cos ⁡ 2 θ \cot \theta \cos ^{2} \theta

Omkar Kulkarni by Omkar Kulkarni , 22, India

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2 solutions

By complementary angles:

sin ⁡ ( π 2 − θ ) = cos ⁡ θ cos ⁡ ( π 2 − θ ) = sin ⁡ θ \sin { \left( \frac { \pi }{ 2 } -\theta \right) } =\cos { \theta } \\ \cos { \left( \frac { \pi }{ 2 } -\theta \right) } =\sin { \theta }

Replacing in the original expression:

cot ⁡ θ − cos ⁡ θ sin ⁡ θ cos ⁡ θ sin ⁡ θ − cos ⁡ θ sin ⁡ θ 1 sin ⁡ θ [ cos ⁡ θ − cos ⁡ θ ( sin ⁡ θ ) 2 ] \cot { \theta } -\cos { \theta } \sin { \theta } \\ \frac { \cos { \theta } }{ \sin { \theta } } -\cos { \theta } \sin { \theta } \\ \frac { 1 }{ \sin { \theta } } \left[ \cos { \theta } -\cos { \theta } { \left( \sin { \theta } \right) }^{ 2 } \right]

By trigonometry identity:

( sin ⁡ θ ) 2 = 1 − ( cos ⁡ θ ) 2 { \left( \sin { \theta } \right) }^{ 2 }=1-{ \left( \cos { \theta } \right) }^{ 2 }

1 sin ⁡ θ [ cos ⁡ θ − cos ⁡ θ − ( cos ⁡ θ ) 3 ] cos ⁡ θ sin ⁡ θ ( cos ⁡ θ ) 2 cot ⁡ θ ( cos ⁡ θ ) 2 \frac { 1 }{ \sin { \theta } } \left[ \cos { \theta } -\cos { \theta } -{ \left( \cos { \theta } \right) }^{ 3 } \right] \\ \frac { \cos { \theta } }{ \sin { \theta } } { \left( \cos { \theta } \right) }^{ 2 }\\ \cot { \theta } { \left( \cos { \theta } \right) }^{ 2 }

2 Helpful 0 Interesting 0 Brilliant 0 Confused
Harsh Soni
Jan 24, 2015

Direct by Hit & Trial Put thetta=π\4 Or try any other value 0f thetta

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