Let cos(α+β)=54 and sin(α−β)=135 , where 0≤α, β≤4π , then tan2α is equal to

tan2α=cos2αsin2α⇒tan2α=1−2sin22αsin2αsin2α=sin[(α+β)+(α−β)]=sin(α+β)cos(α−β)+sin(α−β)cos(α+β)=53×1312+135×54=6536+6520=6556cos2α= 1−2sin22α=1−2(6556)2=6533tan2α=65336556=3356

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Let (+)={4}{5} and (-)={5}{13} , where 0, {}{4} , then 2 i

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

\frac{56}{33}

\frac{19}{12}

\frac{20}{7}

\frac{25}{16}

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Lucas Discussed

Let

cos (α + β) = \frac{4}{5}

and

sin (α - β) = \frac{5}{13}

, where

0 \leq α, β \leq \frac{π}{4}

, then

tan 2 α

is equal to

9 mins ago

Discuss this question LIVE

9 mins ago

Text solutionVerified

tan 2 α = \frac{sin 2 α}{cos 2 α}

\Rightarrow tan 2 α = \frac{sin 2 α}{1 - 2 sin ^{2} 2 α}

sin 2 α = s in [(α + β) + (α - β)] = sin (α + β) cos (α - β) + sin (α - β) cos (α + β)

= \frac{3}{5} \times \frac{12}{13} + \frac{5}{13} \times \frac{4}{5} = \frac{36}{65} + \frac{20}{65} = \frac{56}{65}

cos 2 α = 1 - 2 sin^{2} 2 α = 1 - 2 (\frac{56}{65})^{2} = \frac{33}{65}

tan 2 α = \frac{\frac{56}{65}}{\frac{33}{65}} = \frac{56}{33}

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Question Text	Let $cos (α + β) = \frac{4}{5}$ and $sin (α - β) = \frac{5}{13}$ , where $0 \leq α, β \leq \frac{π}{4}$ , then $tan 2 α$ is equal to
Answer Type	Text solution:1
Upvotes	150

Let cos(α+β)=54​ and sin(α−β)=135​ , where 0≤α, β≤4π​ , then tan2α is equal to

Text solutionVerified

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