Question

tanA/(1+secA)-tanA/(1-secA)=2cosecA
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Answer 1

tanA/(1 + secA) - tanA/(1 - secA) = 2cosecA

Taking L.H.S.

= tanA/(1 + secA) - tanA/(1 - secA)

= tanA [(1 - secA) - (1 + secA)]/[(1 + secA)(1 - secA)]

Used identity: (a + b)(a - b) = a² - b²

= tanA (1 - secA - 1 - secA)/(1 - sec²A)

= tanA (-2 secA)/(1 - sec²A)

We know that, sec²A - 1 = tan²A

= tanA (-2 secA)/(-tan²A)

= 2 secA/tanA

Also, secA = 1/cosA and tanA = sinA/cosA

= 2 (1/cosA)/(sinA/cosA)

= 2/sinA

And 1/sinA = 1/cosecA

= 2 cosecA

L.H.S. = R.H.S.

Hence, proved

Answer 2

Answer:-

We have to prove:

$\sf \dfrac{ \tan \: A}{1 + \sec \:A } - \dfrac{ \tan \: A }{1 - \sec \: A } = 2 \: \csc \: A$

Taking LCM in LHS we get,

$\sf \implies \dfrac{\tan A (1 - \sec A) - tan A (1 + \sec A)}{(1 + \sec A)(1 - \sec A)} = 2 \csc A$

Using the formula (a + b)(a - b) = a² - b² in LHS we get,

$\sf \implies \dfrac{\tan A - \tan A \sec A - tan A - \tan A \sec A}{1^2 - {\sec}^{2} A} = 2 \csc A$

Writing tan A as sin A / Cos A ,sec A as 1/Cos A in LHS we get,

$\sf \implies \dfrac{ - 2 \times\dfrac{ \sin A }{\cos A} \times \dfrac {1}{\cos A} }{1 - \dfrac{1}{\cos^2 A} }= 2 \csc A$

$\sf \implies \dfrac{ \dfrac{- 2\sin A}{\cos^2 A}}{\dfrac{\cos^2A - 1}{\cos^2 A}} = 2 \csc A$

$\sf \implies \dfrac{- 2 \sin A}{\cos ^2 A} \times \dfrac{\cos^2 A}{\cos^2 A - 1} = 2 \csc A$

We know that,

sin² A + cos² A = 1

→ cos² A - 1 = - sin² A

Hence,

$\sf \implies \dfrac{ - 2\sin A}{\cos^2 A} \times \dfrac{\cos^2 A}{- \sin ^2 A} = 2 \csc A$

$\sf \implies 2 \bigg(\dfrac{1}{\sin A}\bigg) = 2 \csc A$

Using 1/sin A = Cosec A in LHS we get,

$\large{\sf{2 \csc A = 2 \csc A}}$

→ LHS = RHS.

Hence ,Proved.