Question

If $\sec a+\tan a=p$, then show that $\sec a-\tan a={\displaystyle \frac{1}{p}}$ . Hence find the value of $\cos a$ and $\sin a$.

Answer 1

Hint- For solving this problem use the basic identities of trigonometry such as ${\sec }^2\theta -{\tan }^2\theta =1$ and ${\sin }^2\theta +{\cos }^2\theta =1$.

Given that:
$\Rightarrow \sec a+\tan a=p$…………………………….. (1)
As we know that
$$\begin{gathered} a^2-b^2=(a+b)(a-b) \\ {\sec }^{2}a-{\tan }^{2}a=1 \end{gathered}$$
Using above formula
$$\begin{gathered} \Rightarrow {\sec }^{2}a-{\tan }^{2}a=1 \\ \Rightarrow (\sec a+\tan a)(\sec a-\tan a)=1 \end{gathered}$$
Using the value given in above equation, we get
$$\begin{gathered} \Rightarrow p(\sec a-\tan a)=1 \\ \Rightarrow \sec a-\tan a={\displaystyle \frac{1}{p}} \end{gathered}$$………………………………… (2)
Hence, we have arrived at our first result.
Now, we have to find out the value of $\cos a$ and $\sin a$.
By adding equation (1) and (2) and further solving , we obtain
$$\begin{gathered} (\sec a+\tan a)+(\sec a-\tan a)=p+{\displaystyle \frac{1}{p}} \\ 2\sec a=p+{\displaystyle \frac{1}{p}} \\ \sec a={\displaystyle \frac{p+{\displaystyle \frac{1}{p}}}{2}} \end{gathered}$$
As we know
 $$\because \cos \theta ={\displaystyle \frac{1}{\sec \theta }}$$
$$\begin{gathered} \cos a={\displaystyle \frac{2}{p+{\displaystyle \frac{1}{p}}}} \\ \cos a={\displaystyle \frac{2p}{p^2+1}} \end{gathered}$$
As we know
 $$\begin{gathered} \because {\sin }^{2}a+{\cos }^{2}a=1 \\ \Rightarrow \sin a=\sqrt{1-{\cos }^{2}a} \end{gathered}$$
Using the value of $\cos a$ in above equation and further solving it, we get
$$\begin{gathered} \sin a=\sqrt{1-{\displaystyle \frac{4p^2}{1+2p^2+p^4}}} \\ =\sqrt{{\displaystyle \frac{1+2p^2+p^4-4p^2}{1+2p^2+p^4}}} \\ =\sqrt{{\displaystyle \frac{1-2p^2+p^4}{1+2p^2+p^4}}} \\ =\sqrt{{\displaystyle \frac{{(1-p^2)}^2}{{(1+p^2)}^2}}}\left[\because {\left(a+b\right)}^2=a^2+b^2+2ab\mathrm{\&}{\left(a-b\right)}^2=a^2+b^2-2ab\right] \\ \sin a={\displaystyle \frac{(1-p^2)}{(1+p^2)}} \end{gathered}$$
So, the values of $\cos a={\displaystyle \frac{2p}{1+p^2}}$ and $\sin a={\displaystyle \frac{(1-p^2)}{(1+p^2)}}$.

Note- Before solving these types of problems you must remember all the trigonometric identities and try to bring all the terms in a single variable. All the same terms will cancel out.