Question

Prove it............​

Answer 1

Answer:

LHS=

$\frac{tanA}{1 + secA} - \frac{tanA}{1 - secA} = tanA(1 - secA - 1 - secA) \\ \frac{tanA( - 2secA)}{1 - {sec}^{2} A} = \frac{2tanA \times secA}{( {sec}^{2} A - 1)} \: \: \: \:[{(a + b)(a - b) = {a}^{2} - {b}^{2}}] \\ \frac{2tanA \times secA}{ {tan}^{2}A } \\ \: { {sec}^{2} A - {tan}^{2} A = 1} \\ \frac{2secA}{tanA} = \frac{2}{sinA} = 2cosecA = rhs$

Answer 2

Answer:

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$\huge\underline\mathrm{To Prove :}$

Step-by-step explanation:

$\frac{tan \: a}{1 + sec \: a} - \frac{tan \: a}{1 - sec \: a} = 2cosec \: a$

Taking L. H. S =

$\frac{tan \: a}{1 + seca} - \frac{tan \: a}{1 - sec \: a}$

Taking LCM of the denominators

$\frac{tan \: a(1 - sec \: a )- tan \: a(1 + sec \: a)}{(1 + sec \: a)(1 - sec \: a)}$

$\frac{(tan \: a - tan \: a \: sec \: a) - (tan \: a + tan \: a \: sec \: a)}{ {(1)}^{2} - {(sec \: a})^{2} }$

(tan A – tan a sec a –tan a – tan a sec a) / 1–sec² a

(–2tan a sec a) / –tan²a

2sec a/ tan a

2×1/cos a / sin a / cos a

2×1/sin a

2cosec a. = R. H. S

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