Question

prove that:

1+(secA -tanA)^2
________________=2 tanA
cosecA(secA -tanA)
​

Answer 1

Answer:

answer for the given problem is given

Answer 2

QUESTION:

$\sf{Prove \ that\ \dfrac{1+(sec\ A-tan \ A)^2}{cosec \ A(sec\ A-tan \ A)}=2tan \ A}$

GIVEN:

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

TO PROVE:

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

FORMULAE:

★ 1 = Sec² A - Tan² A

★ (A - B)² = (A - B)(A - B)

★ A² - B² = (A + B)(A - B)

★ Sec A = 1 / Cos A

★ Cosec A = 1 / Sin A

PROOF:

$\sf{ Take \ \dfrac{1+(sec\ A-tan \ A)^2}{cosec \ A(sec\ A-tan \ A)}\ as \ L.H.S}$

$\sf{Take \ 2tan \ A\ as \ R.H.S}$

1 = Sec² A - Tan² A

$\sf{=\dfrac{sec^2\ A - tan^2 A+(sec\ A-tan \ A)^2}{cosec \ A(sec\ A-tan \ A)}}$

$\bf{Take \ sec\ A - tan \ A\ as\ common}$

$\sf{=\dfrac{sec\ A - tan \ A\ (sec\ A + tan A+sec\ A-tan \ A)}{cosec \ A(sec\ A-tan \ A)}}$

$\sf{=\dfrac{sec\ A + tan A+sec\ A-tan \ A}{cosec \ A}}$

$\sf{=\dfrac{ 2sec\ A}{cosec \ A}}$

$\sf{=\dfrac{ \dfrac{2}{cos\ A}}{\dfrac{1}{sin\ A}}}$

$\sf{=\dfrac{2sin\ A}{cos\ A}}$

$\boxed{\gray{\bold{\large{\dfrac{2sin\ A}{cos\ A}= 2 tan \ A}}}}$

L.H.S = R.H.S

$\sf{ \gray{ \dfrac{1+(sec\ A-tan \ A)^2}{cosec \ A(sec\ A-tan \ A)}=2tan \ A}}$

HENCE PROVED.