Question

How do you prove ${{\tan }^{2}}\left( \dfrac{x}{2}+\dfrac{\pi }{4} \right)=\dfrac{1+\sin x}{1-\sin x}$?

Answer 1

The functions sine, cosine and tangent of an angle are sometimes referred to as the primary or basic trigonometric functions. Trigonometric identities are the equations involving the trigonometric functions that are true for every value of the variables involved. These identities are true for right angled triangles. So, the Pythagorean identity of sine function is ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$.

Complete step by step answer:
As per the given question, we need to prove the given trigonometric expression using trigonometric identities and algebraic formulae. Here, we are given the expression ${{\tan }^{2}}\left( \dfrac{x}{2}+\dfrac{\pi }{4} \right)=\dfrac{1+\sin x}{1-\sin x}$ which we need to verify whether it is correct or not by solving one side of the equation.

Now we consider the right-hand side of the equation that is $\dfrac{1+\sin x}{1-\sin x}$. Now we multiply this with $1+\sin x$ in both numerator and denominator. Then the equation will be

& \Rightarrow \dfrac{1+\sin x}{1-\sin x}\times \left( \dfrac{1+\sin x}{1+\sin x} \right) \\\ & \Rightarrow \dfrac{{{\left( 1+\sin x \right)}^{2}}}{1-{{\sin }^{2}}x} \\\ \end{aligned}$$ We know that, from Pythagorean identity $${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$$. We can rewrite the above equation as $$\begin{aligned} & \Rightarrow \dfrac{{{\left( 1+\sin x \right)}^{2}}}{{{\cos }^{2}}x} \\\ & \Rightarrow {{\left( \dfrac{1+\sin x}{\cos x} \right)}^{2}} \\\ \end{aligned}$$ Let us assume $$t=\tan \dfrac{x}{2}$$ then $$\sin x$$will be $$\dfrac{2t}{1+{{t}^{2}}}$$ and $$\cos x$$ will be $$\dfrac{1-{{t}^{2}}}{1+{{t}^{2}}}$$. On substituting these values in above equation, we get $$\begin{aligned} & \Rightarrow {{\left( \dfrac{1+\dfrac{2t}{1+{{t}^{2}}}}{\dfrac{1-{{t}^{2}}}{1+{{t}^{2}}}} \right)}^{2}} \\\ & \Rightarrow {{\left( \dfrac{1+{{t}^{2}}+2t}{1-{{t}^{2}}} \right)}^{2}} \\\ \end{aligned}$$ $$\Rightarrow {{\left( \dfrac{{{\left( 1+t \right)}^{2}}}{1-{{t}^{2}}} \right)}^{2}}={{\left( \dfrac{{{\left( 1+t \right)}^{2}}}{\left( 1-t \right)\left( 1+t \right)} \right)}^{2}}={{\left( \dfrac{\left( 1+t \right)}{\left( 1-t \right)} \right)}^{2}}$$ Now again we substitute $$t=\tan \dfrac{x}{2}$$ then the equation becomes $$\Rightarrow {{\left( \dfrac{\left( 1+\tan \dfrac{x}{2} \right)}{\left( 1-\tan \dfrac{x}{2} \right)} \right)}^{2}}$$ We know that $$\tan \dfrac{\pi }{4}=1$$. On substituting this value in the above equation. we get $$\Rightarrow {{\left( \dfrac{\left( \tan \dfrac{\pi }{4}+\tan \dfrac{x}{2} \right)}{\left( 1-\tan \dfrac{\pi }{4}\tan \dfrac{x}{2} \right)} \right)}^{2}}$$ We know that $$\tan \left( a+b \right)=\dfrac{\tan a+\tan b}{1-\tan a\tan b}$$. On comparing the above equation with this formula. we can rewrite the equation as $$\Rightarrow {{\left( \tan \left( \dfrac{x}{2}+\dfrac{\pi }{4} \right) \right)}^{2}}$$ $$\Rightarrow {{\tan }^{2}}\left( \dfrac{x}{2}+\dfrac{\pi }{4} \right)$$ which is equal to left hand side of the equation. **Therefore, this means that $${{\tan }^{2}}\left( \dfrac{x}{2}+\dfrac{\pi }{4} \right)=\dfrac{1+\sin x}{1-\sin x}$$. Both the left-hand side and right-hand side are equal. Hence verified.** **Note:** In order to solve such types of questions, we need to have enough knowledge over trigonometric functions and identities. We also need to know the algebraic formulae to simplify the expressions. We must avoid calculation mistakes to get the expected answers.