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If $\sec \theta=\frac{17}{8}$ then prove that $\frac{3-4 \sin ^{2} \theta}{4 \cos ^{2} \theta-3}=\frac{3-\tan ^{2} \theta}{1-3 \tan ^{2} \theta}$.


It is given that $\sec \theta=\frac{17}{8}$.

Let us consider a right $\triangle \mathrm{ABC}$ right angled at $\mathrm{B}$ and $\angle C=\theta$.

We know that $\cos \theta=\frac{1}{\sec \theta}=\frac{8}{17}=\frac{B C}{A C}$

So, if BC = 8k, then AC = 17k, where k is a positive number.
Using Pythagoras theorem, we have:
AC2 = AB2 + BC2
⇒ AB2 = AC2 -"> BC2 = (17k)2 -"> (8k)2
⇒ AB2 = 289k2 -"> 64k2 = 225k2
⇒ AB = 15k.

Now, $\tan \theta=\frac{A B}{B C}=\frac{15}{8}$ and $\sin \theta=\frac{A B}{A C}=\frac{15 k}{17 k}=\frac{15}{17}$

The given expression is $\frac{3-4 \sin ^{2} \theta}{4 \cos ^{2} \theta-3}=\frac{3-\tan ^{2} \theta}{1-3 \tan ^{2} \theta}$.

Substituting the values in the above expression, we get:

LHS $=\frac{3-4\left(\frac{15}{17}\right)^{2}}{4\left(\frac{8}{17}\right)^{2}-3}$






Hence proved.


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