Question

If $sec A=\frac{17}{8}$, verify that $\frac{3-4sin^{2}A }{4cos^{2}A-3 } =\frac{3-tan^{2}A }{1-3tan^{2}A }$

Answer 1

SOLUTION :  

Given: sec A = 17/8

sec A = 17/8 = H / B = AC/AB

Base  = 8 and Hypotenuse = 17

In right angled ΔABC, by using Pythagoras theorem

AC² = AB² + BC²

17² = 8² + BC²

289 = 64  + BC²

BC²  = 289 - 64

BC² = 225

BC = √225

Perpendicular (BC) = 15

Now, sin A = P/H = BC/AC

sin A = 15/17

cos A = B/H = AB/AC

cos A = 8/17

tan A = P/B = BC/AB

tan A = 15/8

CALCULATION of VERIFY that is in the ATTACHMENT.

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Answer 2

hello...

sec A = 17/8

sec A = 17/8 = H / B = AC/AB

Base  = 8

Hypotenuse = 17


AC² = AB² + BC²

17² = 8² + BC²

289 = 64  + BC²

BC²  = 289 - 64

BC² = 225

BC = √225
= 15

sin A = P/H = BC/AC

sin A = 15/17

cos A = B/H = AB/AC

cos A = 8/17

tan A = P/B
= BC/AB

tan A = 15/8


thank you....