If sec A=178, verify that 3−4 sin2 A4 cos2 A−3=3−tan2 A1−3 tan2 A.
Question:

If sec $A=\frac{17}{8}$, verify that $\frac{34 \sin ^{2} A}{4 \cos ^{2} A-3}=\frac{3-\tan ^{2} A}{1-3 \tan ^{2} A}$

Solution:

Given:

$\sec A=\frac{17}{8}$…..(1)

To verify:

$\frac{3-4 \sin ^{2} A}{4 \cos ^{2} A-3}=\frac{3-\tan ^{2} A}{1-3 \tan ^{2} A}$………… (2)

Now we know that $\sec A=\frac{1}{\cos A}$

Therefore $\cos A=\frac{1}{\sec A}$

Now, by substituting the value of $\sec A$ from equation (1)

We get,

$\cos A=\frac{1}{\frac{17}{8}}$

$=\frac{8}{17}$

Therefore,

$\cos A=\frac{8}{17}$…..(3)

Now, we know the following trigonometric identity

$\cos ^{2} A+\sin ^{2} A=1$

Therefore,

$\sin ^{2} A=1-\cos ^{2} A$

Now by substituting the value of $\cos A$ from equation (3)

We get,

$\sin ^{2} A=1-\left(\frac{8}{17}\right)^{2}$

$=1-\frac{(8)^{2}}{(17)^{2}}$

$=1-\frac{64}{289}$

Now by taking L.C.M

We get,

$\sin ^{2} A=\frac{289-64}{289}$

$=\frac{225}{289}$

Now, by taking square root on both sides

We get,

$\sin A=\sqrt{\frac{225}{289}}$

$=\frac{\sqrt{225}}{\sqrt{289}}$

$=\frac{15}{17}$

Therefore,

$\sin A=\frac{15}{17}$….(4)

Now, we know that $\tan A=\frac{\sin A}{\cos A}$

Now by substituting the value of $\cos A$ and $\sin A$ from equation (3) and (4) respectively

We get,

$\tan A=\frac{\frac{15}{17}}{\frac{8}{17}}$

$=\frac{15}{17} \times \frac{17}{8}$

$=\frac{15}{8}$

Therefor

$\tan A=\frac{15}{8} \ldots \ldots$(5)

Now from the expression of equation (2)

L.H.S $=\frac{3-4 \sin ^{2} A}{4 \cos ^{2} A-3}$

Now by substituting the value of $\cos A$ and $\sin A$ from equation (3) and (4)

We get,

L.H.S $=\frac{3-4\left(\frac{15}{17}\right)^{2}}{4\left(\frac{8}{17}\right)^{2}-3}$

Therefore,

L.H.S $=\frac{3-4\left(\frac{225}{289}\right)}{4\left(\frac{64}{289}\right)-3}$

$=\frac{3-\frac{900}{289}}{\frac{256}{289}-3}$

Now by taking L.C.M of both numerator and denominator

We get,

L.H.S $=\frac{\frac{3 \times 289}{1 \times 289}-\frac{900}{289}}{\frac{256}{289}-\frac{3 \times 289}{1 \times 289}}$

$=\frac{\frac{867}{289}-\frac{900}{289}}{\frac{256}{289}-\frac{867}{289}}$

$=\frac{\frac{867-900}{289}}{\frac{256-867}{289}}$

$=\frac{\frac{-33}{289}}{\frac{-611}{289}}$

$=\frac{33}{611}$

Therefore,

$\frac{3-4 \sin ^{2} A}{4 \cos ^{2} A-3}=\frac{33}{611}$///(6)

Now from the expression of equation (2)

R.H.S $=\frac{3-\tan ^{2} A}{1-3 \tan ^{2} A}$

Now by substituting the value of $\tan A$ from equation (5)

We get,

$R . H . S .=\frac{3-\left(\frac{15}{8}\right)^{2}}{1-3\left(\frac{15}{8}\right)^{2}}$

$=\frac{3-\frac{225}{64}}{1-\frac{3 \times 225}{64}}$

Now by taking L.C.M

We get,

R.H.S $=\frac{\frac{3 \times 64}{1 \times 64}-\frac{225}{64}}{\frac{64-675}{64}}$

$=\frac{\frac{192}{64}-\frac{225}{64}}{\frac{-611}{64}}$

$=\frac{\frac{192-225}{64}}{\frac{-611}{64}}$

Therefore

R.H.S $=\frac{\frac{-33}{64}}{\frac{-611}{64}}$

Therefore,

$\frac{3-\tan ^{2} A}{1-3 \tan ^{2} A}=\frac{33}{611}$

Now by comparing equation (6) and (7)

We get,

$\frac{3-4 \sin ^{2} A}{4 \cos ^{2} A-3}=\frac{3-\tan ^{2} A}{1-3 \tan ^{2} A}$