Question

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

Answer 1

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

Let us consider a right $△$ABC right angled at B and $\angle C=\theta$.
We know that cos $\theta$ = $\frac{1}{sec\theta }=\frac{8}{17}=\frac{BC}{AC}$

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 $\theta$ = $\frac{AB}{BC}=\frac{15}{8}$ and sin $\theta$ = $\frac{AB}{AC}=\frac{15k}{17k}=\frac{15}{17}$

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:
$$\begin{gathered} \mathrm{LHS}=\frac{3-4{\left({\displaystyle \frac{15}{17}}\right)}^2}{4{\left({\displaystyle \frac{8}{17}}\right)}^2-3} \\ =\frac{3-{\displaystyle \frac{900}{289}}}{{\displaystyle \frac{256}{289}}-3} \\ =\frac{867-900}{256-867}=\frac{-33}{-611}=\frac{33}{611} \end{gathered}$$

$$\begin{gathered} \mathrm{RHS}=\frac{3-{\left({\displaystyle \frac{15}{8}}\right)}^2}{1-3{\left({\displaystyle \frac{15}{8}}\right)}^2} \\ =\frac{3-{\displaystyle \frac{225}{64}}}{1-{\displaystyle \frac{675}{64}}} \\ =\frac{192-225}{64-675}=\frac{-33}{-611}=\frac{33}{611} \end{gathered}$$

∴ LHS = RHS
Hence proved.