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Question

# If α and β are angles in the first quadrant, and tanα=17, sinβ=1√10, then using the formula sin(A+B)=sinAcosB+cosAsinB, find the value of (α+2β).

A
0
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B
45
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C
60
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D
90
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Solution

## The correct option is B 45∘we know that, sin2β=2tanβ1+tan2βcos2β=1−tan2β1+tan2βalso we know that, 1+tan2α=sec2α⇒1+(17)2=sec2α⇒5049⇒cos2α=4950 ⇒cosα=75√2sinα=√1−4950=15√2also, we know that, sinβ=1√10⇒cosβ=√1−110=3√10∴ tanβ=sinβcosβ=13Now, sin(α+2β)=sinαcos2β+cosαsin2β=15√2×1−tan2β1+tan2β+75√2×2tanβ1+tan2β=15√2[1−191+19+7×231+19]=15√2[810+7×610]=15√2[5010]=15√2×5=1√2⇒sin(α+2β)=1√2⇒(α+2β)=45o

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