Let cosα+β=4/5 and let sinα-β=5/13, where 0≤α,β≤(π)/4, then tan2α=

Correct answer: (C) 56/33. 0≤α,β≤(π)/4; α+β≤(π)/2 cos(α+β)=4/5⇒tan(α+β)=3/4 sin(α-β)=5/13⇒α-β∈θ_1 and tan(α-β)=5/12 tan2α=tan(α+β+α-β) =(tan(α+β)+tan(α-β))/(1-tan(α+β)tan(α-β)) =(3/4+5/12)/(1-3/4×512)=56/33

Q77 / 125 · BITSAT 2019 · Mathematics

Let $\cos (α + β) = \frac{4}{5}$ and let $\sin (α - β) = \frac{5}{13}$ , where $0 \leq α, β \leq \frac{π}{4}$ , then $\tan 2 α =$

A
$\frac{20}{7}$
B
$\frac{25}{16}$
Correct answer
C
$\frac{56}{33}$
D
$\frac{19}{2}$

Solution

$0 \leq α, β \leq \frac{π}{4}; α + β \leq \frac{π}{2}$

$\cos (α + β) = 4 / 5 \Rightarrow \tan (α + β) = 3 / 4$

$\sin (α - β) = 5 / 13 \Rightarrow α - β \in θ_{1}$

and $\tan (α - β) = 5 / 12$

$\tan 2 α = \tan (α + β + α - β)$

$= \frac{\tan (α + β) + \tan (α - β)}{1 - \tan (α + β) \tan (α - β)}$

$= \frac{3 / 4 + 5 / 12}{1 - 3 / 4 \times \frac{5}{12}} = \frac{56}{33}$

Hence, the correct answer is option: 'C'.

Let cosα+β=45 and let sinα-β=513, where 0≤α,β≤π4, then tan2α=

Solution

More Trigonometric Ratios & Identities Questions

Let $\cos (α + β) = \frac{4}{5}$ and let $\sin (α - β) = \frac{5}{13}$ , where $0 \leq α, β \leq \frac{π}{4}$ , then $\tan 2 α =$