If sec A=54, verify that 3 sin A−4 sin3 A4 cos3 A−3 cos A=3 tan A−tan3 A1−3 tan2 A.

Solution:

Given:

$\sec A=\frac{5}{4}$....(1)

To verify:

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

$\cos A=\frac{1}{\frac{5}{4}}$

$=\frac{4}{5}$

Therefore,

$\cos A=\frac{4}{5}$...(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,

$\sin ^{2} A=1-\left(\frac{4}{5}\right)^{2}$

$=1-\frac{(4)^{2}}{(5)^{2}}$

$=1-\frac{16}{25}$

Now by taking L.C.M

We get,

$\sin ^{2} A=\frac{25-16}{25}$

$=\frac{9}{25}$

Now, by taking square root on both sides

We get,

$\sin A=\sqrt{\frac{9}{25}}$

$=\frac{\sqrt{9}}{\sqrt{25}}$

$=\frac{3}{5}$

Therefore,

$\sin A=\frac{3}{5}$.....(4)

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

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

We get,

$\tan A=\frac{\frac{3}{5}}{\frac{4}{5}}$

$=\frac{3}{5} \times \frac{5}{4}$

$=\frac{3}{4}$

Therefore

$\tan A=\frac{3}{4}$....(5)

Now from the expression of equation (2)

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

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

 

We get,

L.H.S $=\frac{3\left(\frac{3}{5}\right)-4\left(\frac{3}{5}\right)^{3}}{4\left(\frac{4}{5}\right)^{3}-3\left(\frac{4}{5}\right)}$

Therefore,

L.H.S $=\frac{\frac{9}{5}-4\left(\frac{27}{125}\right)}{4\left(\frac{64}{125}\right)-\frac{12}{5}}$

$=\frac{\frac{9}{5}-\frac{108}{125}}{\frac{256}{125}-\frac{12}{5}}$

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

We get,

L.H.S $=\frac{\frac{9 \times 25}{5 \times 25}-\frac{108}{125}}{\frac{256}{125}-\frac{12 \times 25}{5 \times 25}}$

$=\frac{\frac{225}{125}-\frac{108}{125}}{\frac{256}{125}-\frac{300}{125}}$

$=\frac{\frac{225-108}{125}}{\frac{256-300}{125}}$

$=\frac{\frac{117}{125}}{\frac{-44}{125}}$

$=\frac{-117}{44}$

$\frac{3 \sin A-4 \sin ^{3} A}{4 \cos ^{3} A-3 \cos A}=\frac{-117}{44}$....… (6)

Now from the expression of equation (2)

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

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

We get,

R.H.S. $=\frac{3\left(\frac{3}{4}\right)-\left(\frac{3}{4}\right)^{3}}{1-3\left(\frac{3}{4}\right)^{2}}$

$=\frac{\frac{9}{4}-\frac{27}{64}}{1-\frac{3 \times 9}{16}}$

Now by taking L.C.M

We get,

$\mathrm{R} . \mathrm{H} . \mathrm{S}=\frac{\frac{9 \times 16}{4 \times 16}-\frac{27}{64}}{\frac{16-27}{16}}$

$=\frac{\frac{144}{64}-\frac{27}{64}}{\frac{-11}{16}}$

$=\frac{\frac{144-27}{64}}{\frac{-11}{16}}$

$=\frac{\frac{117}{64}}{\frac{-11}{16}}$

$=\frac{117}{64} \times \frac{16}{-11}$

Now, $16 \times 4=64$

Therefore,

R.H.S $=\frac{117}{4} \times \frac{1}{-11}$

$=\frac{117 \times 1}{4 \times-11}$

$=\frac{117}{-44}$

$=\frac{-117}{44}$

Therefore,

$\frac{3 \tan A-\tan ^{3} A}{1-3 \tan ^{2} A}=\frac{-117}{44}$.....(7)

Now by comparing equation (6) and (7)

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