Solution : The general solution of \(tan \theta\) = \(tan \alpha\) is given by \(\theta\) = \(n\pi + \alpha\), n \(\in\) Z. Proof : We have, \(tan \theta\) = \(tan \alpha\) \(\implies\) \(sin \theta\over cos \theta\) = \(sin \alpha\over cos \alpha\) \(\implies\) \(sin \theta cos \alpha\) – \(cos \theta sin \alpha\) = 0 \(\implies\) \(sin …
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Solution : The general solution of \(cos \theta\) = \(cos \alpha\) is given by \(\theta\) = \(2n\pi \pm \alpha\), n \(\in\) Z. Proof : We have, \(cos \theta\) = \(cos \alpha\) \(\implies\) \(cos \theta\) – \(cos \alpha\) = 0 \(\implies\) -\(2 sin ({\theta + \alpha\over 2}) sin({\theta – \alpha\over 2})\) = 0 \(\implies\) \(sin ({\theta …
What is the General Solution of \(cos \theta\) = \(cos \alpha\) ? Read More »
Solution : The general solution of \(sin \theta\) = \(sin \alpha\) is given by \(\theta\) = \(n\pi + (-1)^n \alpha\), n \(\in\) Z. Proof : We have, \(sin \theta\) = \(sin \alpha\) \(\implies\) \(sin \theta\) – \(sin \alpha\) = 0 \(\implies\) \(2 sin ({\theta – \alpha\over 2}) cos({\theta + \alpha\over 2})\) = 0 \(\implies\) \(sin …
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Solution : We have, \(\sqrt{3} cos \theta\) + \(sin \theta\) = \(\sqrt{2}\) ………….(i) This is of the form \(a cos\theta\) + \(b sin \theta\) = c, where a = \(\sqrt{3}\), b = 1 and c = \(\sqrt{2}\). Let a = \(r cos\alpha\) and b = \(r sin\alpha\). Then, \(\sqrt{3}\) = …
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Solution : The general solution of \(sin^2 \theta\) = \(sin^2 \alpha\) is given by \(\theta\) = \(n\pi \pm \alpha\), n \(\in\) Z. Proof : We have, \(sin^2 \theta\) =\(sin^2 \alpha\) \(\implies\) \(2 sin^2 \theta\) =\(2 sin^2 \alpha\) \(\implies\) \(1 – cos 2\theta\) = \(1 – cos 2\alpha\) \(\implies\) \(cos 2\theta\) = \(cos 2\alpha\) \(\implies\) \(2\theta\) …
What is the General Solution of \(sin^2 \theta\) =\(sin^2 \alpha\) ? Read More »
Solution : The general solution of \(cos^2 \theta\) = \(cos^2 \alpha\) is given by \(\theta\) = \(n\pi \pm \alpha\), n \(\in\) Z. Proof : We have, \(cos^2 \theta\) =\(cos^2 \alpha\) \(\implies\) \(2 cos^2 \theta\) =\(2 cos^2 \alpha\) \(\implies\) \(1 + cos 2\theta\) = \(1 + cos 2\alpha\) \(\implies\) \(cos 2\theta\) = \(cos 2\alpha\) \(\implies\) \(2\theta\) …
What is the General Solution of \(cos^2 \theta\) =\(cos^2 \alpha\) ? Read More »
Solution : The general solution of \(tan^2 \theta\) = \(tan^2 \alpha\) is given by \(\theta\) = \(n\pi \pm \alpha\), n \(\in\) Z. Proof : We have, \(tan^2 \theta\) =\(tan^2 \alpha\) \(\implies\) \(1 – tan^2\theta\over 1 + tan^2 \theta\) =\(1 – tan^2\alpha\over 1 + tan^2 \alpha\) \(\implies\) \(cos 2\theta\) = \(cos 2\alpha\) \(\implies\) \(2\theta\) = \(2n\pi …
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Solution : The general solution of \(cot \theta\) = 0 is given by \(\theta\) = \((2n + 1){\pi\over 2}\), n \(\in\) Z. Proof : We have, \(cot \theta\) = \(OM\over PM\) \(\therefore\) \(cot \theta\) = 0 \(\implies\) \(OM\over PM\) = 0 \(\implies\) OM = 0 \(\implies\) OP coincides with OY or OY’ \(\implies\) \(\theta\) = …
What is the General Solution of \(Cot \theta\) = 0 ? Read More »
Solution : The general solution of \(cos \theta\) = 0 is given by \(\theta\) = \((2n + 1){\pi\over 2}\), n \(\in\) Z. Proof : We have, \(cos \theta\) = \(PM\over OP\) \(\therefore\) \(cos \theta\) = 0 \(\implies\) \(OM\over OP\) = 0 \(\implies\) OM = 0 \(\implies\) OP coincides with OY or OY’ \(\implies\) \(\theta\) = …
What is the General Solution of \(Cos \theta\) = 0 ? Read More »
Solution : The general solution of \(tan \theta\) = 0 is given by \(\theta\) = \(n\pi\), n \(\in\) Z. Proof : We have, \(tan \theta\) = \(PM\over OM\) \(\therefore\) \(tan \theta\) = 0 \(\implies\) \(PM\over OM\) = 0 \(\implies\) PM = 0 \(\implies\) OP coincides with OX or OX’ \(\implies\) \(\theta\) = 0, \(\pi\), \(2\pi\), …
What is the General Solution of \(Tan \theta\) = 0 ? Read More »
