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If secA=5/4 then tanA/1+tansquareâ€‹ A is a 25/12 b 16/25 c 12/25 d 9/16

Answers (1)

We know that

$\\sec^2A-tan^2A=1 \\Given \ sec A = 5/4\\sec^2A=25/16 \\There fore \\tan^2A=sec^A-1=\frac{25}{16}-1=\frac{9}{16}\\tanA=\sqrt{\frac{9}{16}}=\frac{3}{4} (taking\ only\ positive \ values)$

Therefore

$\frac{tanA}{1+tan^2A}=\frac{tanA}{sec^A}=\frac{\frac{3}{4}}{\frac{25}{16}}=\frac{12}{25}$

Posted by

Safeer PP

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