\( \operatorname{Sec} A+\operatorname{Tan} A=(a+1) /(a-1), \operatorna

1.	\( \operatorname{Sec} A+\operatorname{Tan} A=(a+1) /(a-1), \operatorname{Cos} A= \)
Answer» sec A + tan A = \(\frac{a+1}{a-1}\)----(1) \(\because\) sec²A - tan²A = 1 ⇒ (sec A + tan A) (sec A - tan A) = 1 ⇒ sec A - tan A = \(\frac1{sec A + tan A}\) = \(\cfrac{1}{\frac{a+1}{a-1}}\) = \(\frac{a-1}{a+1}\) ⇒ sec A - tan A = \(\frac{a-1}{a+1}\)----(2) By adding equations (1) & (2), we get 2 sec A = \(\frac{a+1}{a-1}\) + \(\frac{a-1}{a+1}\) = \(\frac{(a+1)^2+(a-1)^2}{(a-1)(a+1)}\) ⇒ 2sec A = \(\frac{2(a^2+1)}{a^2-1}\) ⇒ sec A = \(\frac{a^2+1}{a^2-1}\) \(\therefore\) cos A = \(\frac1{sec A}=\frac{a^2-1}{a^2+1}\)

\( \operatorname{Sec} A+\operatorname{Tan} A=(a+1) /(a-1), \operatorname{Cos} A= \)

Answer»

sec A + tan A = \(\frac{a+1}{a-1}\)----(1)

\(\because\) sec²A - tan²A = 1

⇒ (sec A + tan A) (sec A - tan A) = 1

⇒ sec A - tan A = \(\frac1{sec A + tan A}\) = \(\cfrac{1}{\frac{a+1}{a-1}}\) = \(\frac{a-1}{a+1}\)

⇒ sec A - tan A = \(\frac{a-1}{a+1}\)----(2)

By adding equations (1) & (2), we get

2 sec A = \(\frac{a+1}{a-1}\) + \(\frac{a-1}{a+1}\) = \(\frac{(a+1)^2+(a-1)^2}{(a-1)(a+1)}\)

⇒ 2sec A = \(\frac{2(a^2+1)}{a^2-1}\)

⇒ sec A = \(\frac{a^2+1}{a^2-1}\)

\(\therefore\) cos A = \(\frac1{sec A}=\frac{a^2-1}{a^2+1}\)

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