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Question: If \[\tan \theta + \sec \theta = x\], then prove that \[\sin \theta = \dfrac{{{x^2} - 1}}{{{x^2} + 1...

If tan⁡θ+sec⁡θ=x\tan \theta + \sec \theta = x, then prove that sin⁡θ=x2−1x2+1\sin \theta = \dfrac{{{x^2} - 1}}{{{x^2} + 1}}?

Explanation

Solution

To solve this question first we first we multiply and divide by sec⁡θ−tan⁡θ\sec \theta - \tan \theta on the left-hand side. Then we use trigonometry to eliminate some parts. And try to make an equation in sec⁡θ−tan⁡θ\sec \theta - \tan \theta . Then from these equations try to find the value of sec⁡θ\sec \theta and tan⁡θ\tan \theta . Then try to express sin⁡θ\sin \theta in terms of sec⁡θ\sec \theta and tan⁡θ\tan \theta . And put the values in the final equation and simplify that in order to get the value of sin⁡θ\sin \theta and that is the final answer.

Complete step-by-step answer:
Given,
tan⁡θ+sec⁡θ=x\tan \theta + \sec \theta = x ……(i)

On reciprocating the given equation.
1tan⁡θ+sec⁡θ=1x\dfrac{1}{{\tan \theta + \sec \theta }} = \dfrac{1}{x}
On multiplying and divide by sec⁡θ−tan⁡θ\sec \theta - \tan \theta in left hand side.
sec⁡θ−tan⁡θ(tan⁡θ+sec⁡θ)(sec⁡θ−tan⁡θ)=1x\dfrac{{\sec \theta - \tan \theta }}{{\left( {\tan \theta + \sec \theta } \right)\left( {\sec \theta - \tan \theta } \right)}} = \dfrac{1}{x}
On further calculating
sec⁡θ−tan⁡θ(sec⁡θ)2−(tan⁡θ)2=1x\dfrac{{\sec \theta - \tan \theta }}{{{{\left( {\sec \theta } \right)}^2} - {{\left( {\tan \theta } \right)}^2}}} = \dfrac{1}{x}
Now on using the identity of trigonometry (sec⁡θ)2−(tan⁡θ)2=1{\left( {\sec \theta } \right)^2} - {\left( {\tan \theta } \right)^2} = 1
sec⁡θ−tan⁡θ1=1x\dfrac{{\sec \theta - \tan \theta }}{1} = \dfrac{1}{x}
sec⁡θ−tan⁡θ=1x\sec \theta - \tan \theta = \dfrac{1}{x} ……(ii)
Adding equation (i) and (ii) we get
2sec⁡θ=x+1x2\sec \theta = x + \dfrac{1}{x}
Subtracting (ii) from (i) we get
2tan⁡θ=x−1x2\tan \theta = x - \dfrac{1}{x}
Dividing last two equations
tanθsecθ=x−1xx+1x\dfrac{{tan\theta }}{{sec\theta }} = \dfrac{{x - \dfrac{1}{x}}}{{x + \dfrac{1}{x}}}
On taking LCM on denominator-
tanθsecθ=x2−1xx2+1x\dfrac{{tan\theta }}{{sec\theta }} = \dfrac{{\dfrac{{{x^2} - 1}}{x}}}{{\dfrac{{{x^2} + 1}}{x}}}
On solving the left hand side and simplifying the right hand side.
sin⁡θ=x2−1x2+1\sin \theta = \dfrac{{{x^2} - 1}}{{{x^2} + 1}}
Hence proved.

Note: We can solve this method by another method. Also in that method we start from the right-hand side and try to prove that equal to the left-hand side. We put the value of the x on the right-hand side and then simplify that and try to get sin⁡θ\sin \theta in terms of x. That is our final answer. But this method is the long method and there are more chances to get the wrong answer and we must commit mistakes. In this method, students are not able to figure out how to make another equation and find the value of sec⁡θ\sec \theta and tan⁡θ\tan \theta .