Trigonometry practice problems
Q1. Evaluate:
$$
2\bigl(\cos^4 60^\circ + \sin^4 60^\circ\bigr)
\;-\;\bigl(\tan^2 60^\circ + \cot^2 45^\circ\bigr)
\;+\;3\,\sec^2 30^\circ.
$$
Answer: 5/4
Q2. Prove that $$ \sqrt{\;\sec^2\theta \;+\;cosec^2\theta\;} \;=\;\tan\theta\;+\;\cot\theta. $$ Q3. If $$ \frac{\sin^4x}{2}\;+\;\frac{\cos^4x}{3}\;=\;\frac15, $$ then which are the correct options: $$ \begin{aligned} (a)\quad &\tan^2x \;=\;\frac{2}{3},\\ (b)\quad &\frac{\sin^8x}{8}\;+\;\frac{\cos^8x}{27}\;=\;\frac{1}{125},\\ (c)\quad &\tan^2x \;=\;\frac{1}{3},\\ (d)\quad &\frac{\sin^8x}{8}\;+\;\frac{\cos^8x}{27}\;=\;\frac{2}{125}. \end{aligned} $$ Answers: (a), (b)
Q4. Prove that $$ \Bigl(1 + \frac{1}{\tan^2\theta}\Bigr)\, \Bigl(1 + \frac{1}{\cot^2\theta}\Bigr) \;=\; \frac{1}{\sin^2\theta \;-\;\sin^4\theta}\,. $$
Answer: 5/4
Q2. Prove that $$ \sqrt{\;\sec^2\theta \;+\;cosec^2\theta\;} \;=\;\tan\theta\;+\;\cot\theta. $$ Q3. If $$ \frac{\sin^4x}{2}\;+\;\frac{\cos^4x}{3}\;=\;\frac15, $$ then which are the correct options: $$ \begin{aligned} (a)\quad &\tan^2x \;=\;\frac{2}{3},\\ (b)\quad &\frac{\sin^8x}{8}\;+\;\frac{\cos^8x}{27}\;=\;\frac{1}{125},\\ (c)\quad &\tan^2x \;=\;\frac{1}{3},\\ (d)\quad &\frac{\sin^8x}{8}\;+\;\frac{\cos^8x}{27}\;=\;\frac{2}{125}. \end{aligned} $$ Answers: (a), (b)
Q4. Prove that $$ \Bigl(1 + \frac{1}{\tan^2\theta}\Bigr)\, \Bigl(1 + \frac{1}{\cot^2\theta}\Bigr) \;=\; \frac{1}{\sin^2\theta \;-\;\sin^4\theta}\,. $$
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