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Help me plz.....m stuck in this question.....kindly explain 37 and 38

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one year ago

```							 Dear student Please ask only one question in one thread,38.) tan 3x – tan 2x = 1 + tan 2x tan3x hence tan 3x – tan2x /(1 + tan 2x tan3x) = 1 tan (3x – 2x) = 1 tan x = 1hence x = pi/4 and 5 pi/4 but 2x = pi/2 and 5pi/2 hence no solution
```
one year ago
```							37.2sec 2α=tan β+cot β=>2(1/cos 2α)=(sin β/cos β)+(cos β/sin β)[∵sec x=1/cos x;tan x=sin x/cos x;cot x=cos x/sin x]=>2(1/cos 2α)=(sin²β+cos²β)/sin β•cos β=>(1/cos 2α)=1/2sin β•cos β[∵sinx+cos²x=1]=>cos 2α=2sin β•cos β=>cos 2α=sin 2β[∵2sin x•cos x=sin 2x]=>cos 2α=cos(π/2-2β)[∵sin x=cos(90°-x)]=>2α=π/2-2β=>2α+2β=π/2=>2(α+β)=π/2∴α+β=π/4
```
one year ago
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