Question

if sin Q=12/13 find the value of cos+tan/sec-tan

Answer 1

Answer:

181/13

Step-by-step explanation:

In a triangle, say PQR, sin Q=12/13

This means ratio of the opposite of Q to hypotenuse is 12:13.

Let opposite side=12x and Hypotenuse=13x.

Then, adjacent side, by Pythagoras' Theorem

$=\sqrt{(13x)^2-(12x)^2}\\=\sqrt{169x^2-144x^2}\\=\sqrt{25x^2}\\=5x$

cos Q= Adjacent/hypotenuse=5x/13x=5/13

tan Q= Opposite/adjacent=12x/5x=12/5

sec Q=1/cos Q = 13/5

So,

$\frac{\cos Q+\tan Q}{\sec Q - \tan Q} \\\\=\frac{\frac{5}{13}+\frac{12}{5}}{\frac{13}{5}-\frac{12}{5}} \\\\=\frac{\frac{5(5)+12(13)}{13*5}}{\frac{1}{5}}\\\\=\frac{25+156}{65}*5\\\\=\frac{181}{13}$

Hope it helps!

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