Question

Root 1-tansquare45degree/1+tan square 45degree

Answer 1

Answer:

hey mate it is your answer is 0.

Answer 2

QUESTION :

√1-tan²45°/1 + tan²45°

ANSWER:

We know that

sec²θ - tan²θ = 1

sec²θ = 1 + tan²θ

tan²θ = sec²θ - 1

Substituting the value of tan²θ & 1 + tan²θ = sec²θ

${ \bf{ \sqrt{ \frac{1 - \frac{ {sin}^{2} \theta}{ {cos}^{2} \theta } }{1 + \frac{ {sin}^{2} \theta }{ {cos}^{2} \theta} } } = \sqrt{ \frac{ \frac{ {cos}^{2} \theta - {sin}^{2} \theta }{ {cos}^{2} \theta} }{ \frac{ {cos}^{2} \theta + {sin}^{2} \theta }{ {cos}^{2} \theta} } } }}$

We know that

sin²θ + cos²θ = 1

${ \bf{ \sqrt{ \frac{ {cos}^{2} \theta - {sin}^{2} \theta }{1} } = \sqrt{ { cos}^{2} \theta - {sin}^{2} \theta } }}$

We know that

cos²θ - sin²θ = cos2θ

${ \bf{ \sqrt{ \frac{1 - { tan }^{2} \theta}{1 + {tan}^{2} \theta} } = \sqrt{cos2 \theta}}}$

Given that

θ = 45°

√cos2(45°

√cos90°

√0 = 0