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

Last updated date: 25th Mar 2023

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Answer

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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

$

{a^2} - {b^2} = (a + b)(a - b) \\

{\sec ^2}a - {\tan ^2}a = 1 \\

$

Using above formula

$

\Rightarrow {\sec ^2}a - {\tan ^2}a = 1 \\

\Rightarrow (\sec a + \tan a)(\sec a - \tan a) = 1 \\

$

Using the value given in above equation, we get

$

\Rightarrow p(\sec a - \tan a) = 1 \\

\Rightarrow \sec a - \tan a = \dfrac{1}{p} \\

$………………………………… (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

$

(\sec a + \tan a) + (\sec a - \tan a) = p + \dfrac{1}{p} \\

2\sec a = p + \dfrac{1}{p} \\

\sec a = \dfrac{{p + \dfrac{1}{p}}}{2} \\

$

As we know

\[\because \cos \theta = \dfrac{1}{{\sec \theta }}\]

$

\cos a = \dfrac{2}{{p + \dfrac{1}{p}}} \\

\cos a = \dfrac{{2p}}{{{p^2} + 1}} \\

$

As we know

$

\because {\sin ^2}a + {\cos ^2}a = 1 \\

\Rightarrow \sin a = \sqrt {1 - {{\cos }^2}a} \\

$

Using the value of $\cos a$ in above equation and further solving it, we get

$

\sin a = \sqrt {1 - \dfrac{{4{p^2}}}{{1 + 2{p^2} + {p^4}}}} \\

= \sqrt {\dfrac{{1 + 2{p^2} + {p^4} - 4{p^2}}}{{1 + 2{p^2} + {p^4}}}} \\

= \sqrt {\dfrac{{1 - 2{p^2} + {p^4}}}{{1 + 2{p^2} + {p^4}}}} \\

= \sqrt {\dfrac{{{{(1 - {p^2})}^2}}}{{{{(1 + {p^2})}^2}}}} {\text{ }}\left[ {\because {{\left( {a + b} \right)}^2} = {a^2} + {b^2} + 2ab\& {{\left( {a - b} \right)}^2} = {a^2} + {b^2} - 2ab} \right] \\

\sin a = \dfrac{{(1 - {p^2})}}{{(1 + {p^2})}} \\

$

So, the values of $\cos a = \dfrac{{2p}}{{1 + {p^2}}}$ and $\sin a = \dfrac{{(1 - {p^2})}}{{(1 + {p^2})}}$.

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.

Given that:

$ \Rightarrow \sec a + \tan a = p$…………………………….. (1)

As we know that

$

{a^2} - {b^2} = (a + b)(a - b) \\

{\sec ^2}a - {\tan ^2}a = 1 \\

$

Using above formula

$

\Rightarrow {\sec ^2}a - {\tan ^2}a = 1 \\

\Rightarrow (\sec a + \tan a)(\sec a - \tan a) = 1 \\

$

Using the value given in above equation, we get

$

\Rightarrow p(\sec a - \tan a) = 1 \\

\Rightarrow \sec a - \tan a = \dfrac{1}{p} \\

$………………………………… (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

$

(\sec a + \tan a) + (\sec a - \tan a) = p + \dfrac{1}{p} \\

2\sec a = p + \dfrac{1}{p} \\

\sec a = \dfrac{{p + \dfrac{1}{p}}}{2} \\

$

As we know

\[\because \cos \theta = \dfrac{1}{{\sec \theta }}\]

$

\cos a = \dfrac{2}{{p + \dfrac{1}{p}}} \\

\cos a = \dfrac{{2p}}{{{p^2} + 1}} \\

$

As we know

$

\because {\sin ^2}a + {\cos ^2}a = 1 \\

\Rightarrow \sin a = \sqrt {1 - {{\cos }^2}a} \\

$

Using the value of $\cos a$ in above equation and further solving it, we get

$

\sin a = \sqrt {1 - \dfrac{{4{p^2}}}{{1 + 2{p^2} + {p^4}}}} \\

= \sqrt {\dfrac{{1 + 2{p^2} + {p^4} - 4{p^2}}}{{1 + 2{p^2} + {p^4}}}} \\

= \sqrt {\dfrac{{1 - 2{p^2} + {p^4}}}{{1 + 2{p^2} + {p^4}}}} \\

= \sqrt {\dfrac{{{{(1 - {p^2})}^2}}}{{{{(1 + {p^2})}^2}}}} {\text{ }}\left[ {\because {{\left( {a + b} \right)}^2} = {a^2} + {b^2} + 2ab\& {{\left( {a - b} \right)}^2} = {a^2} + {b^2} - 2ab} \right] \\

\sin a = \dfrac{{(1 - {p^2})}}{{(1 + {p^2})}} \\

$

So, the values of $\cos a = \dfrac{{2p}}{{1 + {p^2}}}$ and $\sin a = \dfrac{{(1 - {p^2})}}{{(1 + {p^2})}}$.

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.

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