Question
If $\sec\theta=\frac{17}{8},$ verify that $\frac{\big(3-4\sin^2\theta\big)}{\big(4\cos^2\theta-3\big)}=\frac{\big(3-\tan^2\theta\big)}{\big(1-3\tan^2\theta\big)}.$
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Answer
Given: $\sec\theta=\frac{17}{8}=\frac{17\text{k}}{8\text{k}}$ Let us draw a $\triangle\text{ABC},$ in which $\angle\text{B}=90^\circ$ and $\angle\text{A}=\theta$
By pythagoras theoram we have $\text{AC}^2 = \text{AB}^2 + \text{BC}^2$ Or $\text{BC}^2 = \text{AC}^2 + \text{AB}^2$ $\therefore\text{BC}^2=\big(17\text{k}\big)^2-\big(8\text{k}\big)^2$ $=289\text{k}^2-64\text{k}^2=225\text{k}^2$ $=\text{BC}=15\text{k}$ $\therefore\ \sin\theta=\frac{\text{AC}}{\text{BC}}=\frac{15{\text{k}}}{17\text{k}}=\frac{15}{17}$ $\cos\theta=\frac{\text{AB}}{\text{AC}}=\frac{8\text{k}}{17\text{k}}=\frac{8}{17}$ $\tan\theta=\frac{15\text{k}}{8\text{k}}=\frac{15}{8}$ $\text{L.H.S.}=\frac{3-4\sin^2\theta}{4\cos^2\theta-3}$ $=\frac{3-4\times\big(\frac{15}{17}\big)^2}{4\times\big(\frac{8}{17}\big)^2-3}$ $=\frac{3-\frac{4\times225}{289}}{4\times\frac{64}{289}-3}$ $=\frac{\frac{3\times289-4\times225}{289}}{\frac{4\times64-3\times289}{289}}$ $=\frac{867-900}{256-867}=\frac{-33}{-611}=\frac{33}{611}$ $\text{R.H.S.}=\frac{3-\tan^2\theta}{1-3\tan^2\theta}=\frac{3-\big(\frac{15}{8}\big)^2}{1-3\times\big(\frac{15}{8}\big)^2}$ $=\frac{3-\frac{225}{64}}{1-3\times\frac{225}{64}}=\frac{\frac{3\times64-225}{64}}{\frac{64-3\times225}{64}}$ $=\frac{192-225}{64-675}=\frac{-33}{-611}=\frac{33}{611}$ Hence, L.H.S = R.H.S.
By pythagoras theoram we have $\text{AC}^2 = \text{AB}^2 + \text{BC}^2$ Or $\text{BC}^2 = \text{AC}^2 + \text{AB}^2$ $\therefore\text{BC}^2=\big(17\text{k}\big)^2-\big(8\text{k}\big)^2$ $=289\text{k}^2-64\text{k}^2=225\text{k}^2$ $=\text{BC}=15\text{k}$ $\therefore\ \sin\theta=\frac{\text{AC}}{\text{BC}}=\frac{15{\text{k}}}{17\text{k}}=\frac{15}{17}$ $\cos\theta=\frac{\text{AB}}{\text{AC}}=\frac{8\text{k}}{17\text{k}}=\frac{8}{17}$ $\tan\theta=\frac{15\text{k}}{8\text{k}}=\frac{15}{8}$ $\text{L.H.S.}=\frac{3-4\sin^2\theta}{4\cos^2\theta-3}$ $=\frac{3-4\times\big(\frac{15}{17}\big)^2}{4\times\big(\frac{8}{17}\big)^2-3}$ $=\frac{3-\frac{4\times225}{289}}{4\times\frac{64}{289}-3}$ $=\frac{\frac{3\times289-4\times225}{289}}{\frac{4\times64-3\times289}{289}}$ $=\frac{867-900}{256-867}=\frac{-33}{-611}=\frac{33}{611}$ $\text{R.H.S.}=\frac{3-\tan^2\theta}{1-3\tan^2\theta}=\frac{3-\big(\frac{15}{8}\big)^2}{1-3\times\big(\frac{15}{8}\big)^2}$ $=\frac{3-\frac{225}{64}}{1-3\times\frac{225}{64}}=\frac{\frac{3\times64-225}{64}}{\frac{64-3\times225}{64}}$ $=\frac{192-225}{64-675}=\frac{-33}{-611}=\frac{33}{611}$ Hence, L.H.S = R.H.S.
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