Trigonometry! #92

Geometry Level 2

Select the correct simplification of sec ⁡ θ − 1 sec ⁡ θ + 1 + sec ⁡ θ + 1 sec ⁡ θ − 1 \sqrt{\frac{\sec \theta - 1}{\sec \theta + 1}} + \sqrt{\frac{\sec \theta + 1}{\sec \theta - 1}}

This problem is part of the set Trigonometry .

2 csc ⁡ θ 2\csc\theta 2 sec ⁡ θ 2\sec\theta 2 sin ⁡ θ sec ⁡ θ \frac{2\sin\theta}{\sqrt{\sec\theta}} 2 cos ⁡ θ 2\cos\theta

Omkar Kulkarni by Omkar Kulkarni , 22, India

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2 solutions

Radhesh Sarma
Feb 7, 2015

2 Helpful 0 Interesting 0 Brilliant 0 Confused

For making your task easier, you could've made t a n θ tan \ \theta in both denominators.

Kunal Verma - 5 years, 11 months ago
Munem Shahriar
Apr 18, 2018

sec ⁡ θ − 1 sec ⁡ θ + 1 + sec ⁡ θ + 1 sec ⁡ θ − 1 = ( sec ⁡ θ − 1 ) ( sec ⁡ θ − 1 ) ( sec ⁡ θ + 1 ) ( sec ⁡ θ − 1 ) + ( sec ⁡ θ + 1 ) ( sec ⁡ θ + 1 ) ( sec ⁡ θ − 1 ) ( sec ⁡ θ + 1 ) = ( sec ⁡ θ − 1 ) 2 sec ⁡ 2 θ − 1 + ( sec ⁡ θ + 1 ) 2 sec ⁡ 2 θ − 1 = sec ⁡ θ − 1 tan ⁡ 2 θ + sec ⁡ θ + 1 tan ⁡ 2 θ [ sec ⁡ 2 θ − 1 = tan ⁡ 2 θ ] = sec ⁡ θ − 1 tan ⁡ θ + sec ⁡ θ + 1 tan ⁡ θ = 2 sec ⁡ θ tan ⁡ θ = 2 cos ⁡ θ × cos ⁡ θ sin ⁡ θ = 2 sin ⁡ θ = 2 1 csc ⁡ θ = 2 csc ⁡ θ \large \begin{aligned} \sqrt{\frac{\sec \theta - 1}{\sec \theta + 1}} + \sqrt{\frac{\sec \theta + 1}{\sec \theta - 1}} & = \sqrt{\frac{(\sec \theta - 1)(\sec \theta - 1)}{(\sec \theta + 1)(\sec \theta - 1)}} + \sqrt{\frac{(\sec \theta +1)(\sec \theta + 1)}{(\sec \theta - 1)(\sec \theta + 1)}} \\ & = \frac{\sqrt{(\sec \theta -1)^2}}{\sqrt{\sec^2 \theta - 1}} + \frac{\sqrt{(\sec \theta + 1)^2}}{\sqrt{\sec^2 \theta - 1}} \\ & = \frac{\sec \theta - 1}{\sqrt{\tan^2 \theta}} + \frac{\sec \theta + 1}{\sqrt{\tan^2 \theta}} ~~~~~~~~~ [\sec^2 \theta - 1 = \tan^2 \theta]\\ & = \frac{\sec \theta - 1}{ \tan \theta} + \frac{\sec \theta + 1}{\tan \theta} \\ & = \frac{2\sec \theta}{\tan \theta} \\ & = \frac 2{\cos \theta} \times \frac{\cos \theta}{\sin \theta} \\ & = \frac 2{\sin \theta} \\ & = \frac 2{\frac 1{\csc \theta}} \\ & = 2 \csc \theta \\ \end{aligned}

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