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Question: If \(\sec a + \tan a = p\), then show that \(\sec a - \tan a = \dfrac{1}{p}\) . Hence find the value...

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

Explanation

Solution

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

Given that:
⇒sec⁡a+tan⁡a=p\Rightarrow \sec a + \tan a = p…………………………….. (1)
As we know that
a2−b2=(a+b)(a−b) sec⁡2a−tan⁡2a=1  {a^2} - {b^2} = (a + b)(a - b) \\\ {\sec ^2}a - {\tan ^2}a = 1 \\\
Using above formula
⇒sec⁡2a−tan⁡2a=1 ⇒(sec⁡a+tan⁡a)(sec⁡a−tan⁡a)=1  \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
⇒p(sec⁡a−tan⁡a)=1 ⇒sec⁡a−tan⁡a=1p  \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\cos a and sin⁡a\sin a.
By adding equation (1) and (2) and further solving , we obtain
(sec⁡a+tan⁡a)+(sec⁡a−tan⁡a)=p+1p 2sec⁡a=p+1p sec⁡a=p+1p2  (\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
∵cos⁡θ=1sec⁡θ\because \cos \theta = \dfrac{1}{{\sec \theta }}
cos⁡a=2p+1p cos⁡a=2pp2+1  \cos a = \dfrac{2}{{p + \dfrac{1}{p}}} \\\ \cos a = \dfrac{{2p}}{{{p^2} + 1}} \\\
As we know
∵sin⁡2a+cos⁡2a=1 ⇒sin⁡a=1−cos⁡2a  \because {\sin ^2}a + {\cos ^2}a = 1 \\\ \Rightarrow \sin a = \sqrt {1 - {{\cos }^2}a} \\\
Using the value of cos⁡a\cos a in above equation and further solving it, we get
sin⁡a=1−4p21+2p2+p4 =1+2p2+p4−4p21+2p2+p4 =1−2p2+p41+2p2+p4 =(1−p2)2(1+p2)2 [∵(a+b)2=a2+b2+2ab&(a−b)2=a2+b2−2ab] sin⁡a=(1−p2)(1+p2)  \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=2p1+p2\cos a = \dfrac{{2p}}{{1 + {p^2}}} and sin⁡a=(1−p2)(1+p2)\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.