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Miscellaneous Exercise 2 · Q56

Q.An expression in tan⁡A\tan A and sec⁡A\sec A is equal to: A) 2 cosec A2\,\text{cosec}\,A B) 2sec⁡A2\sec A C) 2sin⁡A2\sin A D) 2cos⁡A2\cos A (the printed source scan is corrupted at the exact arrangement of the expression whose value is being asked).

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Concept understanding — Fundamental Trigonometric Identities

A trigonometric identity is an equation in trigonometric functions that holds for every admissible value of the angle, not just for special ones. The three fundamental identities all descend from a single geometric fact — that a point (cosθ, sinθ) always lies on the unit circle, so x2+y2=1x^2+y^2=1 becomes

sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1

Dividing this identity through by cos⁡2θ\cos^2\theta gives tan⁡2θ+1=sec⁡2θ\tan^2\theta + 1 = \sec^2\theta, i.e. 1+tan⁡2θ=sec⁡2θ1+\tan^2\theta=\sec^2\theta (valid wherever cosθ ≠ 0); dividing instead by sin⁡2θ\sin^2\theta gives 1+cot⁡2θ=cosec2θ1+\cot^2\theta=\text{cosec}^2\theta (valid wherever sinθ ≠ 0). These three identities are the basic toolkit for simplifying trigonometric expressions: they let you rewrite sec⁡2θ−1\sec^2\theta - 1 as tan⁡2θ\tan^2\theta, combine fractions with denominators like 1−sin⁡θ1-\sin\theta and 1+sin⁡θ1+\sin\theta (whose product is cos⁡2θ\cos^2\theta), find one trigonometric function once another is known (up to a sign fixed by the quadrant), turn an equation like 2sin⁡2θ+7cos⁡θ=52\sin^2\theta+7\cos\theta=5 into a solvable quadratic in a single ratio, and prove that two differently-written expressions are actually equal. Nearly every worked example and identity-proof in this chapter reduces, at some step, to applying one of these three relations.

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