Question

Secø+tanø=x then find sinø

Answer 1

Given: secθ + tanθ = x -->(i)

We know that: sec²θ = 1 + tan²θ

=> sec²θ - tan²θ = 1

=>(secθ + tanθ)(secθ - tanθ)= 1

=> secθ - tanθ = $\frac{1}{secθ + tanθ}$

=> secθ - tanθ = $\frac{1}{x}$ --->(ii)

Adding (i) and (ii):

=> 2secθ = x + $\frac{1}{x}$

=>2secθ = $\frac{x^2 + 1}{x}$

=> $\frac{2}{cosθ}$ = $\frac{x^2 + 1}{x}$

=> cosθ = $\frac{2x}{x^2 + 1}$

Using the identity: Sin²θ + Cos²θ = 1

=> sinθ = $\sqrt{1 - cos^2θ}$

=> sinθ = $\sqrt{1 - (\frac{2x}{x^2 + 1})^2}$

=> sinθ = $\sqrt\frac{x^4 + 1 + 2x^2 - 4x^2}{(x^2 + 1)^2}$

=> sinθ = $\sqrt\frac{x^4 + 1 - 2x^2}{(x^2 + 1)^2}$

=> sinθ = $\sqrt\frac{(x^2 - 1)^2}{(x^2 + 1)^2}$

=> $\bold\color{red}\fbox{sinθ = \frac{x^2 - 1}{x^2 + 1}}$