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The value of $(2cos^{2}\theta-1)\left(\frac{1 + tan\theta}{1 - tan\theta}+\frac{1 - tan\theta}{1 + tan\theta}\right)$

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The value of $(2cos^{2}\theta-1)\left(\frac{1 + tan\theta}{1 - tan\theta}+\frac{1 - tan\theta}{1 + tan\theta}\right)$
1). 4
2). 1
3). 3
4). 2

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2 answers

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Answered by on | Votes 0 |
2 : - is correct hence option 4

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Answered by on | Votes 2 |

$(\frac{1+tan\theta}{1-tan\theta}+\frac{1-tan\theta}{1+tan\theta})$

= (2 cos2$\theta$ – 1)

$(2(\frac{1+tan^{2}\theta}{1-tan^{2}\theta}))$
=$\frac{2sec^{2}\theta(2cos^{2}\theta-1}{1-\frac{sin^{2}\theta}{cos^{2}\theta}}$ = 2.




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