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BITSAT 2019
Exam
BITSAT
Year
2019
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

BITSAT 2019 · Mathematics · Trigonometric Ratios & Identities

Let (α + β) = (4)/(5) and let (α - β) = (5)/(13) , where 0 ≤ α, β ≤...

Let cos⁡(α+β)=45\cos \left(\alpha + \beta\right) = \dfrac{4}{5} and let sin⁡(α−β)=513\sin \left(\alpha - \beta\right) = \dfrac{5}{13}, where 0≤α,β≤π40 \le \alpha, \beta \le \dfrac{\pi}{4}, then tan⁡2α=\tan 2\alpha =

Options

  1. A207\dfrac{20}{7}
  2. B2516\dfrac{25}{16}
  3. C5633\dfrac{56}{33}
  4. D192\dfrac{19}{2}

Correct answer

C. 5633\dfrac{56}{33}

Step-by-step solution

0≤α,β≤π4;α+β≤π20 \le \alpha, \beta \le \dfrac{\pi}{4}; \alpha + \beta \le \dfrac{\pi}{2}

cos⁡(α+β)=4/5⇒tan⁡(α+β)=3/4\cos ( \alpha + \beta ) = 4 / 5 ⇒ \tan ( \alpha + \beta ) = 3 / 4

sin⁡(α−β)=5/13⇒α−β∈θ1\sin ( \alpha - \beta ) = 5 / 13 ⇒ \alpha - \beta \in {θ}_{1}

and tan⁡(α−β)=5/12\tan ( \alpha - \beta ) = 5 / 12

tan⁡2α=tan⁡(α+β+α−β)\tan 2\alpha = \tan ( \alpha + \beta + \alpha - \beta )

=tan⁡(α+β)+tan⁡(α−β)1−tan⁡(α+β)tan⁡(α−β)= \dfrac{\tan ( \alpha + \beta ) + \tan ( \alpha - \beta )}{1 - \tan ( \alpha + \beta ) \tan ( \alpha - \beta )}

=3/4+5/121−3/4×512=5633= \dfrac{3 / 4 + 5 / 12}{1 - 3 / 4 \times \dfrac{5}{12}} = \dfrac{56}{33}

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