A geometry problem by Anuj Shikarkhane

Geometry Level 2

Evaluate :

1 − sin ⁡ θ 1 + sin ⁡ θ \large\sqrt{\dfrac{1-\sin\theta}{1+\sin\theta}}

tan ⁡ θ − sec ⁡ θ \tan\theta - \sec\theta sec ⁡ θ − tan ⁡ θ \sec \theta- \tan\theta sec ⁡ θ − cot ⁡ θ \sec\theta - \cot\theta

Anuj Shikarkhane by Anuj Shikarkhane , 19, India

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2 solutions

Marta Reece
Jun 5, 2017

4 Helpful 0 Interesting 1 Brilliant 0 Confused

Great solution! I am impressed by its beauty ;) One question: How can we derivate (1 - sin^2) from (1 + sin) * (1 - sin)?

Alex Waldherr - 1 year, 10 months ago
Maximos Stratis
Jun 5, 2017

We take 1 − sin ⁡ θ 1 + sin ⁡ θ \sqrt{\frac{1-\sin{\theta}}{1+\sin{\theta}}} and multiply numerator and denominator by 1 − sin ⁡ θ 1-\sin{\theta}
( 1 − sin ⁡ θ ) 2 1 − sin ⁡ 2 θ = \sqrt{\frac{(1-\sin{\theta})^{2}}{1-\sin^{2}{\theta}}}=
( 1 − sin ⁡ θ ) 2 cos ⁡ 2 θ = \sqrt{\frac{(1-\sin{\theta})^{2}}{\cos^{2}{\theta}}}=
1 − sin ⁡ θ cos ⁡ θ = \frac{1-\sin{\theta}}{\cos{\theta}}=
1 cos ⁡ θ − sin ⁡ θ cos ⁡ θ = \frac{1}{\cos{\theta}}-\frac{\sin{\theta}}{\cos{\theta}}=
sec ⁡ θ − tan ⁡ θ \boxed{\sec{\theta}-\tan{\theta}}


1 Helpful 0 Interesting 0 Brilliant 0 Confused

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