The expression is equivalent to

MATHEMATICS · Single correct · CBSE · Mathematics · Introduction to Trigonometry · Introduction to Trigonometry

Problem

The expression tan⁡θ+sec⁡θsec⁡θ−tan⁡θ\displaystyle \frac{\tan\theta+\sec\theta}{\sec\theta-\tan\theta} is equivalent to (a) 1+cos⁡θ1−cos⁡θ\dfrac{1+\cos\theta}{1-\cos\theta} (b) 1+sin⁡θ1−sin⁡θ\dfrac{1+\sin\theta}{1-\sin\theta} (c) 1−sin⁡θ1+sin⁡θ\dfrac{1-\sin\theta}{1+\sin\theta} (d) tan⁡2θ+1\tan^{2}\theta+1

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