Question

If $\sec x+\tan x=k$ then prove that $\sin x=\dfrac{{{k}^{2}}-1}{{{k}^{2}}+1}$.

Answer 1

Hint: In this question, we need to prove $\sin x=\dfrac{{{k}^{2}}-1}{{{k}^{2}}+1}$ if $\sec x+\tan x=k$. For this, we will use trigonometric properties of secx and tanx to convert them into sinx and cosx. After that we will convert cosx into sinx and solve the formed equation in terms of sinx to find the value of sinx in terms of k. We will use the following properties.
\[\begin{align}
  & \left( i \right)\tan x=\dfrac{\sin x}{\cos x} \\
 & \left( ii \right)\sec x=\dfrac{1}{\cos x} \\
 & \left( iii \right){{\sin }^{2}}x+{{\cos }^{2}}x=1\Rightarrow 1-{{\sin }^{2}}x={{\cos }^{2}}x\Rightarrow \cos x=\sqrt{1-{{\sin }^{2}}x} \\
 & \left( iv \right){{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right) \\
\end{align}\]

Complete step by step answer:
Here we are given the equation as $\sec x+\tan x=k$.
We know that secx is reciprocal of cosx. So $\sec x=\dfrac{1}{\cos x}$. Also, $\tan x=\dfrac{\sin x}{\cos x}$ we get $\dfrac{1}{\cos x}+\dfrac{\sin x}{\cos x}=k\Rightarrow \dfrac{1+\sin x}{\cos x}=k$.
We know that ${{\sin }^{2}}x+{{\cos }^{2}}x=1$.
Rearranging we can say ${{\cos }^{2}}x=1-{{\sin }^{2}}x$.
Taking square root on both sides we get $\cos x=\sqrt{1-{{\sin }^{2}}x}$.
Putting this value in the above equation we get $\dfrac{1+\sin x}{\sqrt{1-{{\sin }^{2}}x}}=k$.
Taking square on both sides we get $\dfrac{{{\left( 1+\sin x \right)}^{2}}}{{{\left( \sqrt{1-{{\sin }^{2}}x} \right)}^{2}}}={{k}^{2}}$.
Simplifying the denominator on left side we get $\dfrac{{{\left( 1+\sin x \right)}^{2}}}{1-{{\sin }^{2}}x}={{k}^{2}}$.
As denominator of left side can be written in the form ${{1}^{2}}-{{\sin }^{2}}x$ and we know ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$. So we get $\dfrac{{{\left( 1+\sin x \right)}^{2}}}{\left( 1+\sin x \right)\left( 1-\sin x \right)}={{k}^{2}}$.
Cancelling one of the $\left( 1+\sin x \right)$ from the numerator and the denominator on the left side we get \[\dfrac{1+\sin x}{1-\sin x}={{k}^{2}}\].
Cross multiplying we get \[1+\sin x={{k}^{2}}\left( 1-\sin x \right)\Rightarrow 1+\sin x={{k}^{2}}-{{k}^{2}}\sin x\].
Taking terms containing sinx on one side and rest on the other side we get $\sin x+{{k}^{2}}\sin x={{k}^{2}}-1$.
Taking sinx common from the left side we get $\sin x\left( 1+{{k}^{2}} \right)=1-{{k}^{2}}$.
Dividing both sides by $1+{{k}^{2}}$ we get $\sin x=\dfrac{1-{{k}^{2}}}{1+{{k}^{2}}}$.
Hence we have proved the required result.

Note: Students should keep in mind all the trigonometric identities before solving this sum. Take care of the signs while applying ${{\sin }^{2}}x+{{\cos }^{2}}x=1$. Take care of the signs while using the identity ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$. Take care of the signs while solving the equation in terms of sinx and k. Students can solve these sums using different methods.