Question

please solve it fast​

Answer 1

Answer:

1

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Answer 2

Step-by-step explanation:

To prove :  $\frac{1+tan\alpha }{1-tan\alpha}$ - $\frac{{1-tan\alpha} }{{1+tan\alpha} }$ = $\frac{4sin\alpha cos\alpha }{1-2sin^2\alpha }$

LHS =  $\frac{1+tan\alpha }{1-tan\alpha}$ - $\frac{{1-tan\alpha} }{{1+tan\alpha} }$

=  $\frac{(1+tan\alpha)^2 - (1-tan\alpha)^2 }{1^2-tan^2\alpha}$

= 4(1)(tanα) / 1-tan²α

= 4* $\frac{\frac{sin\alpha}{cos\alpha} }{1- (\frac{sin^2\alpha}{cos^2\alpha}) }$         [ ∵ $tan\alpha = \frac{sin\alpha}{cos\alpha}$ ]

= 4* $\frac{\frac{sin\alpha}{cos\alpha} }{(\frac{cos^2\alpha - sin^2\alpha}{cos^2\alpha}) }$

= 4 * $\frac{sin\alpha}{cos\alpha}$ * $\frac{cos^2\alpha}{cos^2\alpha - sin^2\alpha}$

= 4 *  $\frac{sin\alpha * cos\alpha}{cos^2\alpha - sin^2\alpha}$

= $\frac{4sin\alpha cos\alpha }{1-2sin^2\alpha }$                 [ ∵ $cos^2\alpha - sin^2\alpha = 1-2sin^2\alpha$ ]

= RHS

=> LHS = RHS

Hence proved