Question

(sec theta- tan theta)²= 1-sin theta/1+sin theta ​

Answer 1

Answer:

LHS=(sec∅ - tan∅)²

=(1/cos∅-sin∅/cos∅)²

=(1-sin∅/cos∅)²

LHS=1+sin²∅-2sin∅/cos²∅

RHS=1-sin∅/1+sin∅

=(1-sin∅)(1-sin∅)/(1+sin∅)(1-sin∅)

=(1-sin∅)²/1-sin²∅

RHS=1+sin²∅-2sin∅/cos²∅

Hence, LHS = RHS

Hence proved

Answer 2

Answer:

Prove that: ( Sec theta - tan theta )²

= 1- sin theta / 1+ sin theta

Now, R.H.S : = 1- sin theta / 1 + sin theta

= (1- sin theta) ( 1- sin theta)

______________________

( 1+ sin theta ) ( 1- sin theta)

= ( 1-sin theta )²

____________............[ since,(a+b) (a-b) = a²-b²]

( 1- sin² theta)

= (1-sin theta )²

____________..........[ Since, 1- sin²A = Cos²A]

( cos ² theta )

= [( 1- sin theta) / (Cos theta)]²...[ since , (a/b)^ m =

a^m/b^m ]

= [( 1/Cos theta ) - (sin theta / Cos theta ) ]²

= ( sec theta - tan theta )²

= L.H.S

Therefore, L.H.S = R.H S

Hence , proved