Question

with appropriate explanations ​

Answer 1

★Heya★

We know that

1 + Tan² x = Sec² x

Sec² x - Tan² x = 1

=>

(-Tan x + Sec x ) × ( Sec x + Tan x ) = 1 = y

=>

y = 1

Answer 2

For all possible values of x,

$= ( \sec(x) + \tan(x) )( \sec(x) - \tan(x) )$

$= \sec {}^{2} ( x ) - \tan {}^{2} ( x )$

$= 1$

So, for θ=15° also,
Y = 1.

Extra:-

Proof for
$\sec {}^{2} ( x ) - \tan {}^{2} ( x ) = 1$

$lhs \: = \frac{1}{ \cos {}^{2} (x) } - \frac{ \sin {}^{2} (x) }{\cos {}^{2} (x) }$

$lhs = \frac{1 - \sin {}^{2} (x) }{ \ \cos {}^{2} (x) }$

$lhs = \frac{ \cos {}^{2} (x) }{ \ \cos {}^{2} (x) }$

$lhs = 1$

$lhs = rhs$