Question

answer fast plz it's urgent

Answer 1

Answer:

Step-by-step explanation:

Q.24: Here x = cosec A + cos A --- ( i )

y = cosec A - cos A --- ( ii )

Adding ( i ) and ( ii )

x + y = cosec A + cos A + cosec A - cos A

x + y = 2 cosec A ----- ( iii )

Subtracting ( ii ) from ( i )

x - y = cosec A + cos A - cosec A + cos A

x - y = 2 cos A ---- ( iv )

Now, we have to prove

$(\frac{2}{x+y})^2  + (\frac{x-y}{2})^2 =1$

LHS

$(\frac{2}{x+y})^2  + (\frac{x-y}{2})^2$

Putting values of (x + y) and (x - y) from eq. ( iii ) and ( iv)

$(\frac{2}{2 cosec A})^2  + (\frac{2cosA}{2})^2$

$\frac{1}{cosec^2A}+cos^2A$

[∵ 1/cosecФ = sinФ ∴ 1/cosec²A = sin²A]

sin² A + cos² A

= 1    [∵ sin²Ф + cos²Ф = 1]

= RHS

Similarly you can do Q.35.