Question

TAN A is equal to 3/4 then cos 2A + sec2A=?​

Answer 1

Given that:-

$tanA=\dfrac{3}{4}$

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To find :-

$cos2A+sec2A$

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we know that :-

$cos2x=\dfrac{1-tan^2x}{1+tan^2x}\\ \\ \\ \\sec2x=\dfrac{1+tan^2x}{1-tan^2x}}$

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thus:-

$tanA=\dfrac{3}{4}\\ \\ \\tan^2A=(\dfrac{3}{4})^2\\ \\ \\ \implies tan^2A=\dfrac{9}{16}$

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Thus putting the values of $tan^2x$:-

$\begin{aligned}cos2A+sec2A&=\dfrac{1-tan^2A}{1+tan^2A}+\dfrac{1+tan^2A}{1-tan^2A}}\\ \\ \\cos2A+sec2A&=\dfrac{(1-tan^2A)^2+(1+tan^2A)^2}{(1-tan^2A)(1+tan^2A)}\end{aligned}$

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$=\dfrac{(1-tan^2A)^2+(1+tan^2A)^2}{1^2-(tan^2A)^2}\\ \\ \\=\dfrac{(1-\dfrac{9}{16})^2+(1+\dfrac{9}{16})^2}{1-(\dfrac{9}{16})^2}\\ \\ \\=\dfrac{674}{247}=2.7287...$

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