Question

Prove that:
(i) $\frac{\sin A+\sin 3A+\sin 5A}{\cos A+\cos 3A+\cos 5A}=\tan 3A$

(ii) $\frac{\cos 3A+2\cos 5A+\cos 7A}{\cos A+2\cos 3A+\cos 5A}=\frac{\cos 5A}{\cos 3A}$

(iii) $\frac{\cos 4A+\cos 3A+\cos 2A}{\sin 4A+\sin 3A+\sin 2A}=\cot 3A$

(iv) $\frac{\sin 3A+\sin 5A+\sin 7A+\sin 9A}{\cos 3A+\cos 5A+\cos 7A+\cos 9A}=\tan 6A$

(v) $\frac{\sin 5A-\sin 7A+\sin 8A-\sin 4A}{\cos 4A+\cos 7A-\cos 5A-\cos 8A}=\cot 6A$

(vi) $\frac{\sin 5A\cos 2A-\sin 6A\cos A}{\sin A\sin 2A-\cos 2A\cos 3A}=\tan A$

(vii) $\frac{\sin 11A\sin A+\sin 7A\sin 3A}{\cos 11A\sin A+\cos 7A\sin 3A}=\tan 8A$

(viii) $\frac{\sin 3A\cos 4A-\sin A\cos 2A}{\sin 4A\sin A+\cos 6A\cos A}=\tan 2A$

(ix) $\frac{\sin A\sin 2A+\sin 3A\sin 6A}{\sin A\cos 2A+\sin 3A\cos 6A}=\tan 5A$

(x) $\frac{\sin A+2\sin 3A+\sin 5A}{\sin 3A+2\sin 5A+\sin 7A}=\frac{\sin 3A}{\sin 5A}$

(xi) $\frac{\sin \left(\mathrm{\theta }+\mathrm{\varphi }\right)-2\sin \mathrm{\theta }+\sin \left(\mathrm{\theta }-\mathrm{\varphi }\right)}{\cos \left(\mathrm{\theta }+\mathrm{\varphi }\right)-2\cos \mathrm{\theta }+\cos \left(\mathrm{\theta }-\mathrm{\varphi }\right)}=\tan \mathrm{\theta }$

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right) \\ \mathrm{Consider}\mathrm{LHS}: \\ \frac{\sin A+\sin 3A+\sin 5A}{\cos A+\cos 3A+\cos 5A} \\ =\frac{\sin A+\sin 5A+\sin 3A}{\cos A+\cos 5A+\cos 3A} \\ =\frac{2\sin \left({\displaystyle \frac{A+5A}{2}}\right)\cos \left({\displaystyle \frac{A-5A}{2}}\right)+\sin 3A}{2\cos \left({\displaystyle \frac{A+5A}{2}}\right)\cos \left({\displaystyle \frac{A-5A}{2}}\right)+\cos 3A} \\ =\frac{2\sin 3A\cos \left(-2A\right)+\sin 3A}{2\cos 3A\cos \left(-2A\right)+\cos 3A} \\ =\frac{2\sin 3A\cos 2A+\sin 3A}{2\cos 3A\cos 2A+\cos 3A} \\ =\frac{\sin 3A\left[2\cos 2A+1\right]}{\cos 3A\left[2\cos 2A+1\right]} \\ =\tan 3A \\ =\mathrm{RHS} \\ \mathrm{Hence},\mathrm{RHS}=\mathrm{LHS}. \end{gathered}$$


$$\begin{gathered} \left(\mathrm{ii}\right) \\ \mathrm{Consider}\mathrm{LHS}: \\ \frac{\cos 3A+2\cos 5A+\cos 7A}{\cos A+2\cos 3A+\cos 5A} \\ =\frac{\cos 3A+\cos 7A+2\cos 5A}{\cos A+\cos 5A+2\cos 3A} \\ =\frac{2\cos \left({\displaystyle \frac{3A+7A}{2}}\right)\cos \left({\displaystyle \frac{3A-7A}{2}}\right)+2\cos 5A}{2\cos \left({\displaystyle \frac{A+5A}{2}}\right)\cos \left({\displaystyle \frac{A-5A}{2}}\right)+2\cos 3A} \\ =\frac{2\cos 5A\cos \left(-2A\right)+2\cos 5A}{2\cos 3A\cos \left(-2A\right)+2\cos 3A} \\ =\frac{2\cos 5A\cos 2A+2\cos 5A}{2\cos 3A\cos 2A+2\cos 3A} \\ =\frac{2\cos 5A\left[\cos 2A+1\right]}{2\cos 3A\left[\cos 2A+1\right]} \\ =\frac{\cos 5A}{\cos 3A} \\ =\mathrm{RHS} \\ \mathrm{Hence},\mathrm{RHS}=\mathrm{LHS}. \end{gathered}$$


$$\begin{gathered} \left(\mathrm{iii}\right) \\ \mathrm{Consider}\mathrm{LHS}: \\ \frac{\cos 4A+\cos 3A+\cos 2A}{\sin 4A+\sin 3A+\sin 2A} \\ =\frac{\cos 4A+\cos 2A+\cos 3A}{\sin 4A+\sin 2A+\sin 3A} \\ =\frac{2\cos \left({\displaystyle \frac{4A+2A}{2}}\right)\cos \left({\displaystyle \frac{4A-2A}{2}}\right)+\cos 3A}{2\sin \left({\displaystyle \frac{4A+2A}{2}}\right)\cos \left({\displaystyle \frac{4A-2A}{2}}\right)+\sin 3A} \\ =\frac{2\cos 3A\cos A+\cos 3A}{2\sin 3A\cos A+\sin 3A} \\ =\frac{\cos 3A\left[2\cos A+1\right]}{\sin 3A\left[2\cos A+1\right]} \\ =\cot 3A \\ =\mathrm{RHS} \\ \mathrm{Hence},\mathrm{RHS}=\mathrm{LHS}. \end{gathered}$$
$$\begin{gathered} \left(\mathrm{iv}\right) \\ \mathrm{Consider}\mathrm{LHS}: \\ \frac{\sin 3A+\sin 5A+\sin 7A+\sin 9A}{\cos 3A+\cos 5A+\cos 7A+\cos 9A} \\ =\frac{\sin 3A+\sin 9A+\sin 5A+\sin 7A}{\cos 3A+\cos 9A+\cos 5A+\sin 7A} \\ =\frac{2\sin \left({\displaystyle \frac{3A+9A}{2}}\right)\cos \left({\displaystyle \frac{3A-9A}{2}}\right)+2\sin \left({\displaystyle \frac{5A+7A}{2}}\right)\cos \left({\displaystyle \frac{5A-7A}{2}}\right)}{2\cos \left({\displaystyle \frac{3A+9A}{2}}\right)\cos \left({\displaystyle \frac{3A-9A}{2}}\right)+2\cos \left({\displaystyle \frac{5A+7A}{2}}\right)\cos \left({\displaystyle \frac{5A-7A}{2}}\right)} \\ =\frac{2\sin 6A\cos \left(-3A\right)+2\sin 6A\cos \left(-A\right)}{2\cos 6A\cos \left(-3A\right)+2\cos 6A\cos \left(-A\right)} \\ =\frac{2\sin 6A\cos 3A+2\sin 6A\cos A}{2\cos 6A\cos 3A+2\cos 6A\cos A} \\ =\frac{2\sin 6A\left[\cos 3A+\cos A\right]}{2\cos 6A\left[\cos 3A+\cos A\right]} \\ =\tan 6A \\ =\mathrm{RHS} \\ \mathrm{Hence},\mathrm{LHS}=\mathrm{RH}\text{S.} \end{gathered}$$


$$\begin{gathered} \left(\mathrm{v}\right) \\ \mathrm{Consider}\mathrm{LHS}: \\ \frac{\sin 5A-\sin 7A+\sin 8A-\sin 4A}{\cos 4A+\cos 7A-\cos 5A-\cos 8A} \\ =\frac{\sin 5A-\sin 7A+\sin 8A-\sin 4A}{\cos 4A-\cos 8A+\cos 7A-\sin 5A} \\ =\frac{2\sin \left({\displaystyle \frac{5A-7A}{2}}\right)\cos \left({\displaystyle \frac{5A+7A}{2}}\right)+2\sin \left({\displaystyle \frac{8A-4A}{2}}\right)\cos \left({\displaystyle \frac{8A+4A}{2}}\right)}{-2\sin \left({\displaystyle \frac{4A+8A}{2}}\right)\sin \left({\displaystyle \frac{4A-8A}{2}}\right)-2\sin \left({\displaystyle \frac{7A+5A}{2}}\right)\sin \left({\displaystyle \frac{7A-5A}{2}}\right)} \\ =\frac{2\sin \left(-A\right)\cos 6A+2\sin 2A\cos 6A}{-2\sin 6A\sin \left(-2A\right)-2\sin 6A\sin A} \\ =\frac{-2\sin A\cos 6A+2\sin 2Ac\mathrm{os}6A}{2\sin 6A\sin 2A-2\sin 6A\sin A} \\ =\frac{2\cos 6A\left[\sin 2A-\sin A\right]}{2\sin 6A\left[\sin 2A-\sin A\right]} \\ =\cot 6A \\ =\mathrm{RHS} \\ \mathrm{Hence},\mathrm{LHS}=\mathrm{RHS}. \end{gathered}$$


$$\begin{gathered} \left(\mathrm{vi}\right) \\ \text{Consider}\mathrm{LHS}: \\ \frac{\sin 5A\cos 2A-\sin 6A\cos A}{\sin A\sin 2A-\cos 2A\cos 3A} \\ \mathrm{Multiplying}\mathrm{numerator}\mathrm{and}\mathrm{denominator}\mathrm{by}2,\mathrm{we}\mathrm{get} \\ =\frac{2\sin 5A\cos 2A-2\sin 6\mathrm{A}\cos A}{2\sin A\sin 2\mathrm{A}-2\cos 2A\cos 3\mathrm{A}} \\ =\frac{\sin \left(5\mathrm{A}+2\mathrm{A}\right)+\sin \left(5\mathrm{A}-2\mathrm{A}\right)-\sin \left(6A+\mathrm{A}\right)-\sin \left(6\mathrm{A}-\mathrm{A}\right)}{\cos \left(\mathrm{A}-2\mathrm{A}\right)+\cos \left(\mathrm{A}+2\mathrm{A}\right)-\cos \left(2\mathrm{A}+3\mathrm{A}\right)-\cos \left(2\mathrm{A}-3\mathrm{A}\right)} \\ =\frac{\sin 7A+\sin 3\mathrm{A}-\sin 7\mathrm{A}-\sin 5\mathrm{A}}{\cos \left(-\mathrm{A}\right)+\cos 3\mathrm{A}-\cos 5\mathrm{A}-\cos \left(-\mathrm{A}\right)} \\ =\frac{\sin 7A+\sin 3\mathrm{A}-\sin 7\mathrm{A}-\sin 5\mathrm{A}}{\cos A+\cos 3\mathrm{A}-\cos 5\mathrm{A}-\mathrm{co}sA} \\ =\frac{\sin 3A-\sin 5A}{\cos 3A-\cos 5A} \\ =\frac{2\sin \left({\displaystyle \frac{3A-5\mathrm{A}}{2}}\right)\cos \left({\displaystyle \frac{3A+5A}{2}}\right)}{-2\cos \left({\displaystyle \frac{3A+5A}{2}}\right)\cos \left({\displaystyle \frac{3A-5A}{2}}\right)} \\ =\frac{\sin \left(-A\right)\cos 4A}{-\cos 4A\cos \left(-A\right)} \\ =\frac{-\sin A\cos 4A}{-\cos 4A\cos A} \\ =\frac{\sin A}{\cos A} \\ =\tan A \\ =\mathrm{RHS} \\ \mathrm{Hence},\mathrm{LHS}=\mathrm{RHS}. \end{gathered}$$



$$\begin{gathered} \left(\mathrm{vii}\right) \\ \mathrm{Consider}\mathrm{LHS}: \\ \frac{\sin 11A\sin A+\sin 7A\sin 3A}{\cos 2A\sin A+\cos 6A\sin 3A} \\ \mathrm{Multiplying}\mathrm{numerator}\mathrm{and}\mathrm{denominator}\mathrm{by}2,\mathrm{we}\mathrm{get} \\ =\frac{2\sin 11\mathrm{A}\sin A+2\sin 7A\sin 3A}{2\cos 11\mathrm{A}\mathrm{si}n\mathrm{A}+2\cos 7\mathrm{A}\sin 3\mathrm{A}} \\ =\frac{\cos \left(11\mathrm{A}-\mathrm{A}\right)-\cos \left(11\mathrm{A}+\mathrm{A}\right)+\cos \left(7\mathrm{A}-3\mathrm{A}\right)-\cos \left(7\mathrm{A}+3\mathrm{A}\right)}{\sin \left(11\mathrm{A}+\mathrm{A}\right)-\sin \left(11\mathrm{A}-\mathrm{A}\right)+\sin \left(7\mathrm{A}+3\mathrm{A}\right)-\sin \left(7\mathrm{A}-3\mathrm{A}\right)} \\ =\frac{\cos 10\mathrm{A}-\cos 12\mathrm{A}+\cos 4\mathrm{A}-\cos 10\mathrm{A}}{\sin 12\mathrm{A}-\sin 10\mathrm{A}+\sin 10\mathrm{A}-\sin 4\mathrm{A}} \\ =\frac{\cos 4A-\cos 12A}{\sin 12A-\sin 4A} \\ =\frac{-2\sin \left({\displaystyle \frac{4A+12\mathrm{A}}{2}}\right)\sin \left({\displaystyle \frac{4A-12A}{2}}\right)}{2\sin \left({\displaystyle \frac{12A-4A}{2}}\right)\cos \left({\displaystyle \frac{12A+4A}{2}}\right)} \\ =\frac{-\sin 8A\sin \left(-4A\right)}{\sin 4A\cos 8A} \\ =\frac{\sin 8A\sin 4A}{\sin 4A\cos 8A} \\ =\tan 8A \\ =\mathrm{RHS} \\ \mathrm{Hence},\mathrm{LHS}=\mathrm{RHS}. \end{gathered}$$

$$\begin{gathered} \left(\mathrm{viii}\right) \\ \mathrm{Consider}\mathrm{LHS}: \\ \frac{\sin 3A\cos 4A-\sin A\cos 2A}{\sin 4As\mathrm{in}A+\cos 6A\cos A} \\ \mathrm{Multiplying}\mathrm{numerator}\mathrm{and}\mathrm{denominator}\mathrm{by}2,\mathrm{we}\mathrm{get} \\ =\frac{2\sin 3\mathrm{A}\cos 4\mathrm{A}-2\sin A\cos 2\mathrm{A}}{2\sin 4\mathrm{A}\sin \mathrm{A}+2\cos 6\mathrm{A}\cos \mathrm{A}} \\ =\frac{\sin \left(3\mathrm{A}+4\mathrm{A}\right)+\sin \left(3\mathrm{A}-4\mathrm{A}\right)-\sin \left(\mathrm{A}+2\mathrm{A}\right)-\sin \left(\mathrm{A}-2\mathrm{A}\right)}{\cos \left(4\mathrm{A}-\mathrm{A}\right)-\cos \left(4\mathrm{A}+\mathrm{A}\right)+\cos \left(6\mathrm{A}+\mathrm{A}\right)+\cos \left(6\mathrm{A}-\mathrm{A}\right)} \\ =\frac{\sin 7A+\sin \left(-\mathrm{A}\right)-\sin 3\mathrm{A}-\sin \left(-\mathrm{A}\right)}{\cos 3\mathrm{A}-\cos 5\mathrm{A}+\cos 7\mathrm{A}+\cos 5\mathrm{A}} \\ =\frac{\sin 7A-\sin A-\sin 3\mathrm{A}+\sin A}{\cos 3\mathrm{A}-\cos 5\mathrm{A}+\cos 7\mathrm{A}+\cos 5\mathrm{A}} \\ =\frac{\sin 7A-\sin 3A}{\cos 3A+\cos 7A} \\ =\frac{2\sin \left({\displaystyle \frac{7A-3\mathrm{A}}{2}}\right)\cos \left({\displaystyle \frac{7A+3A}{2}}\right)}{2\cos \left({\displaystyle \frac{3A+7A}{2}}\right)\cos \left({\displaystyle \frac{3A-7A}{2}}\right)} \\ =\frac{\sin 2A\cos 5A}{\cos 5A\cos \left(-2A\right)} \\ =\frac{\sin 2A\cos 5A}{\cos 5A\cos 2A} \\ =\tan 2A \\ =\mathrm{RHS} \\ \mathrm{Hence},\mathrm{LHS}=\mathrm{RHS}. \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ix}\right) \\ \mathrm{Consider}\mathrm{LHS}: \\ \frac{\sin A\sin 2A+\sin 3A\sin 6A}{\sin A\cos 2A+\sin 3A\cos 6A} \\ \mathrm{Multiplying}\mathrm{numerator}\mathrm{and}\mathrm{denominator}\mathrm{by}2,\mathrm{we}\mathrm{get} \\ =\frac{2\sin \mathrm{A}\sin 2\mathrm{A}+2\sin 3\mathrm{A}\sin 6\mathrm{A}}{2\sin A\cos 2\mathrm{A}+2\sin 3\mathrm{A}\cos 6\mathrm{A}} \\ =\frac{\cos \left(\mathrm{A}-2\mathrm{A}\right)-\cos \left(\mathrm{A}+2\mathrm{A}\right)+\cos \left(3\mathrm{A}-6\mathrm{A}\right)-\cos \left(3\mathrm{A}+6\mathrm{A}\right)}{\sin \left(\mathrm{A}+2\mathrm{A}\right)+\sin \left(\mathrm{A}-2\mathrm{A}\right)+\sin \left(3\mathrm{A}+6\mathrm{A}\right)+\sin \left(3\mathrm{A}-6\mathrm{A}\right)} \\ =\frac{\cos \left(-\mathrm{A}\right)-\cos 3\mathrm{A}+\cos \left(-3\mathrm{A}\right)-\cos 9\mathrm{A}}{\sin 3\mathrm{A}\sin \left(-A\right)+\sin 9\mathrm{A}+\sin \left(-3\mathrm{A}\right)} \\ =\frac{\cos A-\cos 3\mathrm{A}+\cos 3\mathrm{A}-\cos 9\mathrm{A}}{\sin 3\mathrm{A}-\sin A+\sin 9\mathrm{A}-\sin 3\mathrm{A}} \\ =\frac{\cos A-\cos 9\mathrm{A}}{\sin 9\mathrm{A}-\sin \mathrm{A}} \\ =\frac{-2\sin \left({\displaystyle \frac{A+9\mathrm{A}}{2}}\right)\sin \left({\displaystyle \frac{A-9A}{2}}\right)}{2\cos \left({\displaystyle \frac{A+9A}{2}}\right)\sin \left({\displaystyle \frac{9A-A}{2}}\right)} \\ =\frac{\sin 5A\cos 4A}{\sin 5A\cos \left(-4A\right)} \\ =\tan 5A \\ =\mathrm{RHS} \\ \mathrm{Hence},\mathrm{LHS}=\mathrm{RHS}. \end{gathered}$$


$$\begin{gathered} \left(\mathrm{x}\right) \\ \mathrm{Consider}\mathrm{LHS}: \\ \frac{\sin A+2\sin 3A+\sin 5A}{\sin 3A+2\sin 5A+\sin 7A} \\ =\frac{\sin A+\sin 5A+2\sin 3A}{\sin 3A+\sin 7A+2\sin 5A} \\ =\frac{2\sin \left({\displaystyle \frac{A+5A}{2}}\right)\cos \left({\displaystyle \frac{A-5A}{2}}\right)+2\sin 3A}{2\sin \left({\displaystyle \frac{3A+7A}{2}}\right)\cos \left({\displaystyle \frac{3A-7A}{2}}\right)+2\sin 5A} \\ =\frac{2\sin 3A\cos \left(-2A\right)+2\sin 3A}{2\sin 5A\cos \left(-2A\right)+2\sin 5A} \\ =\frac{2\sin 3A\cos 2A+2\sin 3A}{2\sin 5A\cos 2A+2\sin 5A} \\ =\frac{2\sin 3A\left[\cos 2A+1\right]}{2\sin 5A\left[\cos 2A+1\right]} \\ =\frac{\sin 3A}{\sin 5A} \\ =\mathrm{RHS} \\ \mathrm{Hence},\mathrm{LHS}=\mathrm{RHS}. \end{gathered}$$

$$\begin{gathered} \left(\mathrm{xi}\right) \\ \mathrm{Consider}\mathrm{LHS}: \\ \frac{\sin \left(\theta +\varphi \right)-2\sin \theta +\sin \left(\theta -\varphi \right)}{\cos \left(\theta +\varphi \right)-2\cos \theta +\cos \left(\theta -\varphi \right)} \\ =\frac{\sin \left(\theta +\varphi \right)+\sin \left(\theta -\varphi \right)-2\sin \theta }{\cos \left(\theta +\varphi \right)+\cos \left(\theta -\varphi \right)-2\cos \theta } \\ =\frac{2\sin \left({\displaystyle \frac{\theta +\varphi +\theta -\varphi }{2}}\right)\cos \left({\displaystyle \frac{\theta +\varphi -\theta +\varphi }{2}}\right)-2\sin \theta }{2\cos \left({\displaystyle \frac{\theta +\varphi +\theta -\varphi }{2}}\right)\cos \left({\displaystyle \frac{\theta +\varphi -\theta +\varphi }{2}}\right)-2\cos \theta } \\ =\frac{2\sin \theta \cos \varphi -2\sin \theta }{2\cos \theta \cos \varphi -2\cos \theta } \\ =\frac{2\sin \theta \left[\cos \varphi -1\right]}{2\cos \theta \left[\cos \varphi -1\right]} \\ =\tan \theta \\ =\mathrm{RHS} \\ \mathrm{Hence},\mathrm{RHS}=\mathrm{LHS}. \end{gathered}$$