Question

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

Answer 1

Solution:-

$\frac{secA=17}{8,}$

show that = $\frac{3-4si{n}^{2}A}{4co{s}^{2}A-3}$ = $\frac{3ta{n}^{2}A}{1-3ta{n}^{2}A.}$

Given sec A = $\frac{17}{8}$

We know, sec A = $\frac{Hypotenuse}{Base}$= $\frac{17}{8}$

So, draw a right angled triangle PQR, right angled at Q such that ∠ PQR = A

Base QR = 8 and Hypotenuse PR = 17

Using Pythagoras Theorem in Δ PQR

PR² = PQ² + QR²

⇒ (17)² = PQ² + (8)²

⇒ 289 = PQ² + 64

⇒ PQ² = 289 - 64

⇒ PQ² = 225

⇒ PQ = 15

L.H.S. = 3-4sin²A/4cos²A-3

⇒ 3-4(15/17)²/4(8/17)² - 3

⇒ 3-4×(225/289)/4×(64/289) - 3

⇒ {(867-900)/289}/{(256-867)/289}

⇒ -33/289 × 289/-611

= 33/611

R.H.S = (3-tan²A)/(1-3tan²A)

⇒ 3-(15/8)²/1-3(15/8)²

⇒ 3 - (255/64)/1 - (675/64)

⇒ -33/64 × 64/-611

⇒ 33\611

So, L.H.S. = R.H.S.

Hence proved