Question

If sec theta + tan theta = x, then prove that
$\bf sin \theta = \dfrac{x^{2} - 1}{x^2 + 1}$

Don't skip a single step

Answer 1

Given:

$\implies\tt{sec\theta + tan\theta= x}$

To prove:

$\implies\tt{sin \theta = \dfrac{x^{2} - 1}{x^2 + 1}}$

Solution:

$\implies\tt{sec\theta + tan\theta= x\:\:(1)}$

On taking reciprocal on both LHS and RHS, we get:

$\implies\tt{ \dfrac{1}{sec\theta + tan\theta}= \dfrac{1}{x}}$

Now, we are going to apply Rationalisation method. We are going to multiply $\tt{sec\theta - tan\theta}$ on both the numerator and denominator in LHS

$\implies\tt{\dfrac{1(sec\theta-tan\theta)}{sec\theta + tan\theta(sec\theta-tan\theta)}= \dfrac{1}{x}}$

At the denominator side, in LHS, we are going to use the identity:

$\green{\boxed{\tt{(a + b)(a - b) = a {}^{2} - b {}^{2}}}}$

$\implies\tt{\dfrac{sec\theta-tan\theta}{sec^2\theta-tan^2\theta}= \dfrac{1}{x}}$

As we know that,

$\implies\tt{tan^2\theta+1 = sec^2\theta}$

$\implies\boxed{\tt{1 = sec^2\theta-tan^2\theta}}$

On applying the identity, we have:

$\implies\tt{\dfrac{sec\theta-tan\theta}{1}= \dfrac{1}{x}}$

$\implies\tt{sec\theta-tan\theta= \dfrac{1}{x}\:\:(2)}$

On adding Equation 1 and 2, we get:

$\implies\tt{sec\theta + tan\theta+sec\theta-tan\theta= x+ \dfrac{1}{x}}$

$\implies\tt{2sec\theta = \dfrac{x^2+1}{x}\:\:(3)}$

On subtracting Equation 2 from 1, we get:

$\implies\tt{sec\theta + tan\theta-(sec\theta-tan\theta)= x- \dfrac{1}{x}}$

$\implies\tt{sec\theta + tan\theta-sec\theta+tan\theta= \dfrac{x^2-1}{x}}$

$\implies\tt{2tan\theta= \dfrac{x^2-1}{x}\:\:(4)}$

On dividing:

$\implies\tt{\dfrac{(4)}{(3)}}$

$\implies\tt{\dfrac{2tan \theta}{2sec \theta} = \dfrac{\dfrac{x {}^{2} - 1}{x}}{ \dfrac{x {}^{2} + 1}{x}}}$

$\boxed{\tt{sin \theta = \dfrac{x^{2} - 1}{x^2 + 1}}}$

Hence, Proved!

Answer 2

$\sf\color{purple}Given:$

$\sf\color{red}To\: prove:$

$\implies\tt{sin \theta = \dfrac{x^{2} - 1}{x^2 + 1}}$

$\sf\color{red}Solution:$

$\implies\tt{sec\theta + tan\theta= x\:\:(1)}$

On taking reciprocal on both LHS and RHS, we get:

$\implies\tt{ \dfrac{1}{sec\theta + tan\theta}= \dfrac{1}{x}}$

Now, we are going to apply Rationalisation method.

We are going to multiply

$\tt{sec\theta - tan\theta}$

$\implies\tt{\dfrac{1(sec\theta-tan\theta)}{sec\theta + tan\theta(sec\theta-tan\theta)}= \dfrac{1}{x}}$

At the denominator side, in LHS, we are going to use the identity:

$\green{\boxed{\tt{(a + b)(a - b) = a {}^{2} - b {}^{2}}}}$

$\implies\tt{\dfrac{sec\theta-tan\theta}{sec^2\theta-tan^2\theta}= \dfrac{1}{x}}$

$\sf\color{red}As\:we \:know\: that$,

$\implies\tt{tan^2\theta+1 = sec^2\theta}$

$\implies\boxed{\tt{1 = sec^2\theta-tan^2\theta}}$

On applying the identity, we have:

$\implies\tt{\dfrac{sec\theta-tan\theta}{1}= \dfrac{1}{x}}$

$\implies\tt{sec\theta-tan\theta= \dfrac{1}{x}\:\:(2)}$

On adding Equation 1 and 2, we get:

$\implies\tt{sec\theta + tan\theta+sec\theta-tan\theta= x+ \dfrac{1}{x}}$

$\implies\tt{2sec\theta = \dfrac{x^2+1}{x}\:\:(3)}$

On subtracting Equation 2 from 1, we get:

$\implies\tt{sec\theta + tan\theta-(sec\theta-tan\theta)= x- \dfrac{1}{x}}$

$\implies\tt{sec\theta + tan\theta-sec\theta+tan\theta= \dfrac{x^2-1}{x}}$

$\implies\tt{2tan\theta= \dfrac{x^2-1}{x}\:\:(4)}$

On dividing:

$\implies\tt{\dfrac{(4)}{(3)}}$

$\implies\tt{\dfrac{2tan \theta}{2sec \theta} = \dfrac{\dfrac{x {}^{2} - 1}{x}}{ \dfrac{x {}^{2} + 1}{x}}}$

$\boxed{\tt{sin \theta = \dfrac{x^{2} - 1}{x^2 + 1}}}$

$\sf\color{red}Hence,\: Proved!$

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