Question

Sec + tan =   Prove that sin = x^2-1/x^2+1
please help ​

Answer 1

Answer:

$\frac{x^2-1}{x^2+1} = sin\theta$

Step-by-step explanation:

$sec\theta + tan\theta =x\\x= \frac{1}{cos\theta} + \frac{sin\theta}{cos\theta} = \frac{1+ sin\theta}{cos\theta} \\x^2 = \frac{(1+ sin\theta)^2}{cos^2\theta} \\x^2 = \frac{(1+ sin\theta)^2}{1-sin^2\theta} \\x^2 = \frac{(1+ sin\theta)^2}{(1+sin\theta)(1-sin\theta)}\\x^2 = \frac{(1+ sin\theta)}{(1-sin\theta)}\\so\\x^2-1 = \frac{(1+ sin\theta)}{(1-sin\theta)}-1 = \frac{1+ sin\theta-1+sin\theta}{(1-sin\theta)}= \frac{(2sin\theta)}{(1-sin\theta)}$$x^2+1 = \frac{(1+ sin\theta)}{(1-sin\theta)}+-1 = \frac{1+ sin\theta+1-sin\theta}{(1-sin\theta)}= \frac{(2)}{(1-sin\theta)}\\\frac{x^2-1}{x^2+1} = sin\theta$

Answer 2

Answer:

2sin/2= sin

Step-by-step explanation:

Given:- x=secθ+tanθ To prove:-

x^2 +1x^2 −1 =sinθ

Proof:-

x=secθ+tanθ⇒x=cosθ1+sinθ

Squaring both sides, we get

⇒x^2 =cos^2 θ(1+sinθ) ^2⇒x^2= 1−sin ^2θ(1+sinθ)^2⇒x^2 = (1+sinθ)(1−sinθ)(1+sinθ) ^2⇒x^2 = 1−sinθ(1+sinθ)

Therefore,

x^2 +1x ^2 −1= 1−sinθ(1+sinθ) +1−sinθ(1+sinθ)−1=1+sinθ+1−sinθ1+sinθ−1+sinθ=2sin =sinθ