Question

Prove that sec A (1- Sin A) (Sec A + tan A)
=1

Answer 1

$\huge\mathfrak\blue{Answer:}$

Given:

• We have been given a trigonometric Function
• $\sf{sec \: A ( 1 - sin \: A )( sec \: A + tan \: A ) = 1 }$

To Prove:

• We have to prove that
• $\sf{sec \: A ( 1 - sin \: A )( sec \: A + tan \: A ) = 1}$

Solution:

We have to prove that

$\sf{sec \: A ( 1 - sin \: A )( sec \: A + tan \: A ) = 1}$

________________________________

$\underline{\large\sf{\orange{Taking \: LHS}}}$

$\implies \sf{sec \: A ( 1 - sin \: A )( sec \: A + tan \: A) }$

Converting all in terms of sine and cosine

$\implies \sf{sec \: A ( 1 - sin \: A )( sec \: A + tan \: A )}$

$\implies \sf{\dfrac{1}{cos \: A} ( 1 - sin \: A )( \dfrac{1}{cos \: A}+ \dfrac{sin \: A }{cos \: A}) }$

Taking LCM

$\implies \sf{\dfrac{1}{cos \: A} ( 1 - sin \: A )( \dfrac{1 + sin \: A}{cos \: A}) }$

$\implies \sf{\dfrac{1}{cos^2 \: A } \times ( 1 - sin \: A )( 1 + sin \: A) }$

_____________________________

Using algebric identities

$\boxed{\sf{\pink{ (a-b)(a+b) = a^2 - b^2 }}}$

$\implies \sf{\dfrac{1}{cos^2 A} ( 1 - sin^2 \: A )}$

$\boxed{\sf{\green{1 - sin^2 \: A = cos^2 \: A }}}$

$\implies \sf{\dfrac{1}{cos^2 A} \times ( cos^2 \: A )}$

$\implies \sf{\dfrac{cos^2 \: A}{cos^2 A}}$

$\implies \sf{\cancel{\dfrac{cos^2 \: A}{cos^2 A}}}$

$\implies \sf{1 \qquad ( RHS ) }$

_______________________________

$\huge\sf{\red{H}\orange{e}\green{n}\pink{c}\blue{e} \: \purple{P} \red{r}\orange{o}\green{v}\pink{e}\blue{d} \: \red{ !!! }}$

$\large\boxed{\sf{\red{sec \: A ( 1 - sin \: A )( sec \: A + tan \: A ) = 1}}}$

__________________________________