Question

(1+cotA-cosecA)(1+tanA+secA)=2​

Answer 1

Given, (1+cot -cosec A) (1+tanA +secA) =2

LHS= (1+cos A/sinA -1/sinA)

=( 1+ sinA/ cos A +1/cos A)

=( sinA + cosA -1 /sinA)

( cos A + sin A +1 / cosA)

By taking LCM of sinA and cosA.

By using the identity,

(a+b) (a- b) =a^2 -b^2.

we get,

= ( sinA +cosA)^2 - (1)^2 ➗ sin A cosA

=sin^2 + cos^2 + 2 sinA cosA. ➖ 1➗ sinA cos A

(as, sin^2 + cos^2 =1)

= 1 ➕2 sinA cos A ➖1

➗ sinA cos A

= 1➕ 2➖1.

= 2 = RHS. [proved]

Thank you!

And is now waiting for more questions to solve.