Solution : We have 1 + \(sin({\pi\over 4} + \theta)\) + \(2cos({\pi\over 4} – \theta)\) = 1 + \(1\over sqrt{2}\)\((cos\theta + cos\theta)\) + \(\sqrt{2}\)\((cos\theta + cos\theta)\) = 1 + \(({1\over \sqrt{2}} + \sqrt{2})\) + \((cos\theta + cos\theta)\) = 1 + \(({1\over \sqrt{2}} + \sqrt{2})\).\(\sqrt{2}cos(\theta – {pi\over 4})\) \(\therefore\) Maximum Value = 1 + \(({1\over \sqrt{2}} …
Solution : The expression = (sin78 – sin42) – (sin66 – sin6) = 2cos(60)sin(18) – 2cos36.sin30 = sin18 – cos36 = \(({\sqrt{5} – 1\over 4})\) – \(({\sqrt{5} + 1\over 4})\) = -\(1\over 2\) Similar Questions Find the maximum value of 1 + \(sin({\pi\over 4} + \theta)\) + \(2cos({\pi\over 4} – \theta)\) If A + B …
Solution : Here, tanx + secx = 2cosx \(\implies\) sinx + 1 = \(2cos^2x\) \(\implies\) \(2sin^2x\) + sinx – 1 = 0 \(\implies\) sinx = \(1\over 2\), -1 But sinx = -1 \(\implies\) x = \(3\pi\over 2\) for which tanx + secx = 2cosx is not defined. Thus, sinx = …
Find the number of solutions of tanx + secx = 2cosx in [0, \(2\pi\)]. Read More »
Solution : Since, \({1\over 6}sin\theta\), \(cos\theta\) and \(tan\theta\) are in G.P. \(\implies\) \(cos^2\theta\) = \({1\over 6}sin\theta\).\(cos\theta\) \(\implies\) \(6cos^3\theta\) + \(cos^2\theta\) – 1 = 0 \(\therefore\) (\(2cos\theta – 1\))(\(3cos^2\theta\) + \(2cos\theta\) + 1) = 0 \(cos\theta\) = \(1\over 2\) (other values are imaginary) \(cos\theta\) = \(cos\pi\over 3\) \(\theta\) = \(2n\pi \pm {\pi\over 3}\), n …
Solution : \(2log{2\over 5}\) + \(3log{25\over 8}\) + \(log{128\over 625}\) = \(log{2^2\over 5^2}\) + \(log({5^2\over 2^3})^3\) + \(log{2^7\over 5^4}\) = \(log({2^2\over 5^2}{5^6\over 2^9}{2^7\over 5^4})\) = log 1 = 0 Similar Questions Solve for x : \(2^{x + 2}\) > \(({1\over 4})^{1\over x}\). Evaluate the given log : \(81^{l\over {log_5 3}}\) + \(27^{log_9 36}\) + \(3^{4\over …
Find the value of \(2log{2\over 5}\) + \(3log{25\over 8}\) – \(log{625\over 128}\). Read More »
Solution : We have \(2^{x + 2}\) > \(2^{2/x}\). Since the base 2 > 1, we have x + 2 > -\(2\over x\) (the sign of the inequality is retained) Now x + 2 + \(2\over x\) \(\implies\) \(x^2 + 2x + 2\over x\) > 0 \(\implies\) \((x + 1)^2 + 1\over x\) > 0 …
Solve for x : \(2^{x + 2}\) > \(({1\over 4})^{1\over x}\). Read More »
Solution : \(81^{log_3 5}\) + \(3^{3log_9 36}\) + \(3^{4log_9 7}\) \(\implies\) \(3^{4log_3 5}\) + \(3^{log_3 {(36)}^{3/2}}\) + \(3^{log_3 {7}^2}\) = 625 + 216 + 49 = 890. Similar Questions Solve for x : \(2^{x + 2}\) > \(({1\over 4})^{1\over x}\). Find the value of \(2log{2\over 5}\) + \(3log{25\over 8}\) – \(log{625\over 128}\). If \(log_a x\) …
Solution : \(\displaystyle{\lim_{x \to \infty}}\)\(x^2 + x + 1\over {3x^2 + 2x – 5}\) It is (\(\infty\over \infty\) form), Put x = \(1\over y\) = \(\displaystyle{\lim_{y \to 0}}\) \(1 + y + y^2\over {3 + 2y – 5y^2}\) = \(1\over 3\) Similar Questions Evaluate : \(\displaystyle{\lim_{x \to 0}}\) \(x^3 cotx\over {1-cosx}\) Evaluate : \(\displaystyle{\lim_{x \to …
Solution : \(\displaystyle{\lim_{x \to 0}}\) \(x^3 cosx\over {sinx(1-cosx)}\) = \(\displaystyle{\lim_{x \to 0}}\) \(x^3 cosx(1 + cosx)\over {sinxsin^2x}\) = \(\displaystyle{\lim_{x \to 0}}\) \({x^3\over sin^3x}.cosx(1 + cosx)\) = 2 Similar Questions Evaluate the limit : \(\displaystyle{\lim_{x \to \infty}}\) \(x^2 + x + 1\over {3x^2 + 2x – 5}\) Evaluate : \(\displaystyle{\lim_{x \to \infty}}\) \(({7x^2+1\over 5x^2-1})^{x^5\over {1-x^3}}\) Evaluate …
Evaluate : \(\displaystyle{\lim_{x \to 0}}\) \(x^3 cotx\over {1-cosx}\) Read More »
Solution : We have, \(tan^{-1}(1)\) + \(cos^{-1}({-1\over 2})\) + \(sin^{-1}({-1\over 2})\) = \(\pi\over 4\) + \(2\pi\over 3\) – \(\pi\over 6\) = \(3\pi\over 4\) Similar Questions Solve the equation : 2\(tan^{-1}({2x+1})\) = \(cos^{-1}x\) Prove that : \(sin^{-1}{12\over 13}\) + \(cot^{-1}{4\over 3}\) + \(tan^{-1}{63\over 16}\) = \(\pi\) Evaluate \(sin^{-1}(sin10)\) Prove that : \(cos^{-1}{12\over 13}\) + \(sin^{-1}{3\over 5}\) …
