Trigono Met Ry

Level 1

( 1 − sin ⁡ x 1 + sin ⁡ x ) 2 = ? \large \left(\frac{1 - \sin x}{1 + \sin x}\right)^2 = ?

sin ⁡ 3 ( 3 0 ∘ − x 4 ) \sin^3 \left(30^\circ - \frac x4 \right) tan ⁡ 4 ( x 2 − 4 5 ∘ ) \tan^4 \left(\frac x2 - 45^\circ \right) cos ⁡ 3 ( 6 0 ∘ − x 8 ) \cos^3\left(60^\circ - \frac x8\right) cot ⁡ 4 ( x 16 − 9 0 ∘ ) \cot^4 \left(\frac x{16} -90^\circ \right)

Ronojoy Paul by Ronojoy Paul , 21, India

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2 solutions

Chew-Seong Cheong
Apr 16, 2018

( 1 − sin ⁡ x 1 + sin ⁡ x ) 2 = ( 1 − 2 tan ⁡ x 2 1 + tan ⁡ 2 x 2 1 + 2 tan ⁡ x 2 1 + tan ⁡ 2 x 2 ) 2 Using half-angle tangent substitution = ( tan ⁡ 2 x 2 − 2 tan ⁡ x 2 + 1 tan ⁡ 2 x 2 + 2 tan ⁡ x 2 + 1 ) 2 = ( ( tan ⁡ x 2 − 1 ) 2 ( tan ⁡ x 2 + 1 ) 2 ) 2 = ( sin ⁡ x 2 cos ⁡ x 2 − 1 sin ⁡ x 2 cos ⁡ x 2 + 1 ) 4 = ( sin ⁡ x 2 − cos ⁡ x 2 sin ⁡ x 2 + cos ⁡ x 2 ) 4 = ( 1 2 sin ⁡ x 2 − 1 2 cos ⁡ x 2 1 2 sin ⁡ x 2 + 1 2 cos ⁡ x 2 ) 4 = ( sin ⁡ ( x 2 − 4 5 ∘ ) cos ⁡ ( x 2 − 4 5 ∘ ) ) 4 = tan ⁡ 4 ( x 2 − 4 5 ∘ ) \begin{aligned} \left(\frac {1-\sin x}{1+\sin x}\right)^2 & = \left(\frac {1-\frac {2\tan \frac x2}{1+\tan^2 \frac x2}}{1+\frac {2\tan \frac x2}{1+\tan^2 \frac x2}}\right)^2 & \small \color{#3D99F6} \text{Using half-angle tangent substitution} \\ & = \left(\frac {\tan^2 \frac x2 - 2\tan \frac x2 + 1}{\tan^2 \frac x2 + 2\tan \frac x2 + 1}\right)^2 \\ & = \left(\frac {\left(\tan \frac x2 - 1\right)^2}{\left(\tan \frac x2 + 1\right)^2}\right)^2 \\ & = \left(\frac {\frac {\sin \frac x2}{\cos \frac x2}-1}{\frac {\sin \frac x2}{\cos \frac x2}+1}\right)^4 \\ & = \left(\frac {\sin \frac x2 - \cos \frac x2}{\sin \frac x2 + \cos \frac x2}\right)^4 \\ & = \left(\frac {\frac 1{\sqrt 2}\sin \frac x2 - \frac 1{\sqrt 2}\cos \frac x2}{\frac 1{\sqrt 2}\sin \frac x2 + \frac 1{\sqrt 2}\cos \frac x2}\right)^4 \\ & = \left(\frac {\sin \left(\frac x2 - 45^\circ\right)}{\cos \left(\frac x2 - 45^\circ\right)} \right)^4 \\ & = \boxed{\tan^4 \left(\frac x2 - 45^\circ\right)} \end{aligned}

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The question stated with a power of 2 and not to the power of 4 ... The first step needs to be reconsulted

Ronojoy Paul - 3 years, 1 month ago

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Sorry, typos.

Chew-Seong Cheong - 3 years, 1 month ago

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It is still 4 not 2.

( 1 − sin ⁡ x 1 + sin ⁡ x ) 4 \left(\dfrac{1-\sin x}{ 1+ \sin x}\right)^{4} .

Naren Bhandari - 3 years, 1 month ago
Ossama Ismail
Jan 3, 2019

Substitute x=90 on both sides.

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