Question

If tan theta +sec theta=x,then show that sin theta = x sqare -1/x square + 1​

Answer 1

Given:-

$\sf{Tan\theta + Sec\theta = x}$

To:-

Show that $\sf{Sin\theta = \dfrac{x^2 - 1}{x^2 + 1}}$

Solution:-

We have,

$\sf{Tan\theta + Sec\theta = x}$

$\sf{:\implies Sec\theta + Tan\theta = x \longrightarrow[i]}$

Now,

Let us find out the value of $\sf{\dfrac{1}{x}}$

$\sf{\because x = Sec\theta + Tan\theta}$

$\sf{\therefore \dfrac{1}{x} = \dfrac{1}{Sec\theta + Tan\theta}}$

By rationalizing the denominator,

= $\sf{\dfrac{1}{Sec\theta + Tan\theta} \times \dfrac{Sec\theta - Tan\theta}{Sec\theta - Tan\theta}}$

= $\sf{\dfrac{Sec\theta - Tan\theta}{Sec^2\theta - Tan^2\theta}\:\:[\because (a+b)(a-b) = a^2 - b^2]}$

= $\sf{\dfrac{Sec\theta - Tan\theta}{1}\:\:[\because Sec^2\theta - Tan^2\theta = 1]}$

We got the value of $\sf{\dfrac{1}{x}}$ as $\sf{Sec\theta - Tan\theta}$

Therefore,

$\sf{\dfrac{1}{x} = Sec\theta - Tan\theta \longrightarrow[ii]}$

Now,

Adding Equation(i) and (ii)

$\sf{x + \dfrac{1}{x} = Sec\theta + Tan\theta + Sec\theta - Tan\theta}$

= $\sf{\dfrac{x^2 + 1}{x} = 2Sec\theta}$

=> $\sf{Sec\theta = \dfrac{x^2 + 1}{2x}}$

=> $\sf{\dfrac{1}{Sec\theta} = \dfrac{1}{\dfrac{x^2 + 1}{2x}}}$

=> $\sf{Cos\theta = \dfrac{2x}{x^2 + 1}}$

We know,

$\sf{Sin\theta = \sqrt{1 - Cos^2\theta}}$

= $\sf{\sqrt{1 - \bigg(\dfrac{2x}{x^2 + 1}\bigg)^2}}$

= $\sf{\sqrt{\dfrac{(x^2 + 1)^2 - 4x^2}{(x^2 + 1)^2}}}$

= $\sf{\sqrt{\dfrac{x^4 + 2x^2 + 1 - 4x^2}{(x^2+1)^2}}}$

= $\sf{\sqrt{\dfrac{x^4 - 2x^2 + 1}{(x^2 + 1)^2}}}$

= $\sf{\sqrt{\dfrac{(x^2)^2 - 2x^2 + 1}{(x^2 + 1)^2}}}$

= $\sf{\because x^2 - 2xy + y^2 = (x-y)^2}$

= $\sf{\sqrt{\dfrac{(x^2 - 1)^2}{(x^2 + 1)^2}}}$

= $\sf{\dfrac{x-1}{x+1}}$

Hence,

$\sf{Sin\theta = \dfrac{x-1}{x+1}}$

(Proved)

______________________________________

Answer 2

$\huge\underbrace\mathfrak \red{ANSWER }$

$\sf{Tan\theta + Sec\theta = x}$

To:-

Show that $\sf{Sin\theta = \dfrac{x^2 - 1}{x^2 + 1}}$

Solution:-

We have,

$\sf{Tan\theta + Sec\theta = x}$

$\sf{:\implies Sec\theta + Tan\theta = x \longrightarrow[i]}$

Now,

Let us find out the value of $\sf{\dfrac{1}{x}}$

$\sf{\because x = Sec\theta + Tan\theta}$

$\sf{\therefore \dfrac{1}{x} = \dfrac{1}{Sec\theta + Tan\theta}}$

By rationalizing the denominator,

= $\sf{\dfrac{1}{Sec\theta + Tan\theta} \times \dfrac{Sec\theta - Tan\theta}{Sec\theta - Tan\theta}}$

= $\sf{\dfrac{Sec\theta - Tan\theta}{Sec^2\theta - Tan^2\theta}\:\:[\because (a+b)(a-b) = a^2 - b^2]}$

= $\sf{\dfrac{Sec\theta - Tan\theta}{1}\:\:[\because Sec^2\theta - Tan^2\theta = 1]}$

We got the value of $\sf{\dfrac{1}{x}}$ as $\sf{Sec\theta - Tan\theta}$

Therefore,

$\sf{\dfrac{1}{x} = Sec\theta - Tan\theta \longrightarrow[ii]}$

Now,

Adding Equation(i) and (ii)

$\sf{x + \dfrac{1}{x} = Sec\theta + Tan\theta + Sec\theta - Tan\theta}$

= $\sf{\dfrac{x^2 + 1}{x} = 2Sec\theta}$

=> $\sf{Sec\theta = \dfrac{x^2 + 1}{2x}}$

=> $\sf{\dfrac{1}{Sec\theta} = \dfrac{1}{\dfrac{x^2 + 1}{2x}}}$

=> $\sf{Cos\theta = \dfrac{2x}{x^2 + 1}}$

We know,

$\sf{Sin\theta = \sqrt{1 - Cos^2\theta}}$

= $\sf{\sqrt{1 - \bigg(\dfrac{2x}{x^2 + 1}\bigg)^2}}$

= $\sf{\sqrt{\dfrac{(x^2 + 1)^2 - 4x^2}{(x^2 + 1)^2}}}$

= $\sf{\sqrt{\dfrac{x^4 + 2x^2 + 1 - 4x^2}{(x^2+1)^2}}}$

= $\sf{\sqrt{\dfrac{x^4 - 2x^2 + 1}{(x^2 + 1)^2}}}$

= $\sf{\sqrt{\dfrac{(x^2)^2 - 2x^2 + 1}{(x^2 + 1)^2}}}$

= $\sf{\because x^2 - 2xy + y^2 = (x-y)^2}$

= $\sf{\sqrt{\dfrac{(x^2 - 1)^2}{(x^2 + 1)^2}}}$

= $\sf{\dfrac{x-1}{x+1}}$

Hence,

$\sf{Sin\theta = \dfrac{x-1}{x+1}}$

(Proved)

HOPE SO IT WILL HELP....

PLEASE MARK IT AS BRAINLIST....