Question

If $\sec \theta =\frac{17}{8}$, verify that $\frac{3-4si{n}^2\theta }{4co{s}^2\theta -3}=\frac{(3-ta{n}^2\theta )}{(1-3ta{n}^2\theta )}$.

Answer 1

$sec\theta =\frac{17}{8}\Rightarrow cos\theta =\frac{8}{17}co{s}^2\theta =\frac{64}{289}\Rightarrow si{n}^2\theta =1-cos²\theta =1-\frac{64}{289}=\frac{225}{289}\Rightarrow ta{n}^2\theta =\frac{si{n}^2\theta }{co{s}^2\theta }=\frac{225}{64}3-4si{n}^2\theta =3-4\times \frac{225}{289}=\frac{-33}{289}4co{s}^2\theta -3=4\times \frac{64}{289}-3=\frac{-611}{289}L.H.S.\Rightarrow \frac{(3-4si{n}^2\theta )}{(4co{s}^2\theta -3)}=\frac{33}{611}And3-ta{n}^2\theta =3-\frac{225}{64}=\frac{-33}{64}1-3ta{n}^2\theta =1-3\times \frac{225}{64}=\frac{-611}{64}R.H.S.\Rightarrow \frac{(3-ta{n}^2\theta )}{(1-ta{n}^2\theta )}=\frac{33}{611}HenceprovedL.H.S=R.H.S\Rightarrow \frac{(3-4si{n}^2\theta )}{(4co{s}^2\theta -3)}=\frac{(3-ta{n}^2\theta )}{(1-3ta{n}^2\theta )}$