Question

prove that sec ^2alpha - sin^2alpha =cos^2alpha + tan^2alpha​

Answer 1

Answer:

answer

As we know that, 1+tan

2

θ=sec

2

θ

So, we can write

cos

2

α−sin

2

α=sec

2

β−1

⇒sec

2

β=cos

2

α−sin

2

α+1

⇒cos

2

β=

cos

2

α−sin

2

α+(sin

2

α+cos

2

α)

1

(as sin

2

α+cos

2

α=1)

⇒cos

2

β=

2cos

2

α

1

also, sin

2

β=1−cos

2

β=1−

2cos

2

α

1

So, cos

2

β−sin

2

β=

2cos

2

α

1

−(1−

2cos

2

α

1

)

=

2cos

2

α

1

−1+

2cos

2

α

1

=

cos

2

α

1

−1=sec

2

α−1=tan

2

α