# Trigonometric Identities

 1.        Prove that in any triangle $\displaystyle ABC,$ (i) $\displaystyle \sin (A+B) = \sin C.$ (ii) $\displaystyle \cos(A+B) + \cos C = 0.$ (iii) $\displaystyle \cos \frac{A+B}{2} = \sin \frac{C}{2}.$ (iv) $\displaystyle \tan \frac{A+B}{2} = \cot \frac{C}{2}.$
Show/Hide Solution
 (i) $\displaystyle \text{Since}\ A+B+C=180{}^\circ ,$ $\displaystyle \begin{array}{l}\therefore A+B=180{}^\circ -C\\\\\therefore \sin (A+B)=\sin (180{}^\circ -C)\\\\\therefore \sin (A+B)=\sin C\end{array}$ (ii) $\displaystyle \text{Similarly, }\cos (A+B)=\cos (180{}^\circ -C)$ $\displaystyle \begin{array}{l}\therefore \cos (A+B)=-\cos C\\\\\therefore \cos (A+B)+\cos C=0\end{array}$ (iii) $\displaystyle \cos \left( {\frac{{A+B}}{2}} \right)=\cos \left( {\frac{{180{}^\circ -C}}{2}} \right)$ $\displaystyle \therefore \ \cos \left( {\frac{{A+B}}{2}} \right)=\cos \left( {90{}^\circ -\frac{C}{2}} \right)$ $\displaystyle \therefore \ \cos \left( {\frac{{A+B}}{2}} \right)=\sin \frac{C}{2}$ (iv) $\displaystyle \tan \left( {\frac{{A+B}}{2}} \right)=\tan \left( {\frac{{180{}^\circ -C}}{2}} \right)$ $\displaystyle \therefore \ \tan \left( {\frac{{A+B}}{2}} \right)=\tan \left( {90{}^\circ -\frac{C}{2}} \right)$ $\displaystyle \therefore \ \tan \left( {\frac{{A+B}}{2}} \right)=\cot \frac{C}{2}$

 2.        In any quadrilateral $\displaystyle ABCD,$ prove that (i) $\displaystyle \sin (A+B) + \sin (C+D)=0.$ (ii) $\displaystyle \cos (A+B) = \cos (C+D).$
Show/Hide Solution
 (i) $\displaystyle \text{In}\ \text{any}\ \text{quadrilateral}\ ABCD,$ $\displaystyle \begin{array}{l}\ \ \ \ A+B+C+D=360{}^\circ \\\\\therefore \ \ A+B=360{}^\circ -(C+D)\\\\\therefore \ \ \sin (A+B)=\sin \left[ {360{}^\circ -(C+D)} \right]\\\\\therefore \ \ \sin (A+B)=-\sin (C+D)\\\\\therefore \ \ \sin (A+B)+\sin (C+D)=0\end{array}$ (ii) $\displaystyle \ \text{Since }A+B+C+D=360{}^\circ ,$ $\displaystyle \begin{array}{l}\ \ \ \ \ \ A+B=360{}^\circ -(C+D)\\\\\therefore \ \ \cos (A+B)=\cos \left[ {360{}^\circ -(C+D)} \right]\\\\\therefore \ \ \sin (A+B)=\cos (C+D)\end{array}$

 3.        If $\displaystyle A, B, C, D$ be the angles of a cyclic quadrilateral, taken in order, prove that (i) $\displaystyle \cos A + \cos B + \cos C + \cos D=0.$ (ii) $\displaystyle \cos (180{}^\circ +A)+\cos (180{}^\circ +B)+\cos (180{}^\circ +C)-\sin (90{}^\circ +D)=0.$
Show/Hide Solution
 (i) $\displaystyle \text{Since }ABCD\ \text{is a cyclic quadrilateral},$ $\displaystyle \begin{array}{l}\ \ \ \ A+C=180{}^\circ \Rightarrow A=180{}^\circ -C\\\\\ \ \ \ B+D=180{}^\circ \Rightarrow B=180{}^\circ -D\\\\\therefore \ \ \cos A=\cos (180{}^\circ -C)\Rightarrow \cos A=-\cos C\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ --(1)\\\\\ \ \ \ \text{Similarly,}\cos B=\cos (180{}^\circ -D)\Rightarrow \cos B=-\cos D--(2)\\\\\ \ \ \ (1)+(2)\Rightarrow \cos A+\cos B=-\cos C-\cos D\end{array}$ (ii) $\displaystyle \ \cos (180{}^\circ +A)+\cos (180{}^\circ +B)+\cos (180{}^\circ +C)-\sin (90{}^\circ +D)$ $\displaystyle \begin{array}{l}\ \ \ \ =-\cos A-\cos B-\cos C-\cos D\\\\ \ \ \ \ =-(\cos A+\cos B+\cos C+\cos D)\\\\ \ \ \ \ =0\ \ \ \ \ \left[ {\because \text{by}\ (\text{i})} \right]\end{array}$

 4.        If $\displaystyle \tan 35{}^\circ =x,$ prove that $\displaystyle \frac{{\tan 145{}^\circ -\tan 125{}^\circ }}{{1+\tan 145{}^\circ \tan 125{}^\circ }}=\frac{{1-{{x}^{2}}}}{{2x}}.$
Show/Hide Solution
 $\displaystyle \begin{array}{l}\ \ \ \ \tan 35{}^\circ =x\\\\\ \ \ \ \tan 145{}^\circ =\tan (180{}^\circ -35{}^\circ )\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-\tan 35{}^\circ \\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-x\\\\\ \ \ \ \tan 125{}^\circ =\tan (90{}^\circ +35{}^\circ )\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-\cot 35{}^\circ \end{array}$ $\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-\frac{1}{{\tan 35{}^\circ }}$ $\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-\frac{1}{x}$ $\displaystyle \ \ \ \ \ \ \frac{{\tan 145{}^\circ -\tan 125{}^\circ }}{{1+\tan 145{}^\circ \tan 125{}^\circ }}$ $\displaystyle \ \ \ \ =\ \ \frac{{-x-\left(\displaystyle {-\frac{1}{x}} \right)}}{{1-x\left(\displaystyle {-\frac{1}{x}} \right)}}$ $\displaystyle \ \ \ \ =\ \ \frac{{\displaystyle \frac{{1-{{x}^{2}}}}{x}}}{2}$ $\displaystyle \ \ \ \ =\ \ \frac{{1-{{x}^{2}}}}{{2x}}$

 5.        Prove that (i) $\displaystyle \sin (A+B)\sin (A-B)={{\sin }^{2}}A-{{\sin }^{2}}B.$ (ii) $\displaystyle \sin (A+B)\sin (A-B)={{\cos }^{2}}B-{{\cos }^{2}}A.$ (iii) $\displaystyle \cos (A+B)\cos (A-B)={{\cos }^{2}}A-{{\sin }^{2}}B.$ (iv) $\displaystyle \cos (A+B)\cos (A-B)={{\cos }^{2}}B-{{\sin }^{2}}A.$
Show/Hide Solution
 (i)$\displaystyle \ \ \ \sin (A+B)\sin (A-B)$ $\displaystyle \begin{array}{l}\ \ \ \ \ =(\sin A\cos B+\cos A\sin B)(\sin A\cos B-\cos A\sin B)\\\\ \ \ \ \ \ ={{\sin }^{2}}A{{\cos }^{2}}B-{{\cos }^{2}}A{{\sin }^{2}}B\\\\ \ \ \ \ \ ={{\sin }^{2}}A(1-{{\sin }^{2}}B)-(1-{{\sin }^{2}}A){{\sin }^{2}}B\\\\ \ \ \ \ \ ={{\sin }^{2}}A-{{\sin }^{2}}A{{\sin }^{2}}B-{{\sin }^{2}}B+{{\sin }^{2}}A{{\sin }^{2}}B\\\\ \ \ \ \ \ ={{\sin }^{2}}A-{{\sin }^{2}}B\end{array}$ (ii)$\displaystyle \ \ \ \sin (A+B)\sin (A-B)$ $\displaystyle \begin{array}{l}\ \ \ \ \ =(\sin A\cos B+\cos A\sin B)(\sin A\cos B-\cos A\sin B)\\\\ \ \ \ \ \ ={{\sin }^{2}}A{{\cos }^{2}}B-{{\cos }^{2}}A{{\sin }^{2}}B\\\\ \ \ \ \ \ =(1-{{\cos }^{2}}A){{\cos }^{2}}B-{{\cos }^{2}}A(1-{{\cos }^{2}}B)\\\\\ \ \ \ \ ={{\cos }^{2}}B-{{\cos }^{2}}A{{\cos }^{2}}B-{{\cos }^{2}}A+{{\cos }^{2}}A{{\cos }^{2}}B\\\\ \ \ \ \ \ ={{\cos }^{2}}A-{{\cos }^{2}}B\end{array}$ (iii)$\displaystyle \ \ \ \cos (A+B)\cos (A-B)$ $\displaystyle \begin{array}{l}\ \ \ \ \ =(\cos A\cos B-\sin A\sin B)(\cos A\cos B+\sin A\sin B)\\\\\ \ \ \ \ \ ={{\cos }^{2}}A{{\cos }^{2}}B-{{\sin }^{2}}A{{\sin }^{2}}B\\\\\ \ \ \ \ \ ={{\cos }^{2}}A(1-{{\sin }^{2}}B)-(1-{{\cos }^{2}}A){{\sin }^{2}}B\\\\\ \ \ \ \ \ ={{\cos }^{2}}A-{{\cos }^{2}}A{{\sin }^{2}}B-{{\sin }^{2}}B+{{\cos }^{2}}A{{\sin }^{2}}B\\\\\ \ \ \ \ \ ={{\cos }^{2}}A-{{\sin }^{2}}B\end{array}$ (iv) $\displaystyle \ \ \ \cos (A+B)\cos (A-B)$ $\displaystyle \begin{array}{l}\ \ \ \ \ \ \ =(\cos A\cos B-\sin A\sin B)(\cos A\cos B+\sin A\sin B)\\\\\ \ \ \ \ \ \ ={{\cos }^{2}}A{{\cos }^{2}}B-{{\sin }^{2}}A{{\sin }^{2}}B\\\\\ \ \ \ \ \ \ =(1-{{\sin }^{2}}A){{\cos }^{2}}B-{{\sin }^{2}}A(1-{{\cos }^{2}}B)\\\\\ \ \ \ \ \ \ ={{\cos }^{2}}B-{{\sin }^{2}}A{{\cos }^{2}}B-{{\sin }^{2}}A+{{\sin }^{2}}A{{\cos }^{2}}B\\\\\ \ \ \ \ \ \ ={{\cos }^{2}}B-{{\sin }^{2}}A\end{array}$

 6.        Prove that $\displaystyle \frac{{\sin (A-B)}}{{\cos A\cos B}}+\frac{{\sin (B-C)}}{{\cos B\cos C}}+\frac{{\sin (C-A)}}{{\cos C\cos A}}=0.$
Show/Hide Solution
 $\displaystyle \ \ \ \ \frac{{\sin (A-B)}}{{\cos A\cos B}}=\frac{{\sin A\cos B-\cos A\sin B}}{{\cos A\cos B}}$ $\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{\sin A\cos B}}{{\cos A\cos B}}-\frac{{\cos A\sin B}}{{\cos A\cos B}}$ $\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\tan A-\tan B$ $\displaystyle \ \ \ \ \frac{{\sin (B-C)}}{{\cos B\cos C}}=\frac{{\sin B\cos C-\cos B\sin C}}{{\cos B\cos C}}$ $\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{\sin B\cos C}}{{\cos B\cos C}}-\frac{{\cos B\sin C}}{{\cos B\cos C}}$ $\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\tan B-\tan C$ $\displaystyle \ \ \ \ \frac{{\sin (C-A)}}{{\cos C\cos A}}=\frac{{\sin C\cos A-\cos C\sin A}}{{\cos C\cos A}}$ $\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{\sin C\cos A}}{{\cos C\cos A}}-\frac{{\cos C\sin A}}{{\cos C\cos A}}$ $\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\tan C-\tan A$ $\displaystyle \therefore \ \ \ \frac{{\sin (A-B)}}{{\cos A\cos B}}+\frac{{\sin (B-C)}}{{\cos B\cos C}}+\frac{{\sin (C-A)}}{{\cos C\cos A}}$ $\displaystyle \ \ =\tan A-\tan B+\tan B-\tan C+\tan C-\tan A$ $\displaystyle \ \ =0$

 7.        Prove that (i) $\displaystyle \frac{{\cos 17{}^\circ +\sin 17{}^\circ }}{{\cos 17{}^\circ -\sin 17{}^\circ }}=\tan 62{}^\circ .$ (ii) $\displaystyle \tan 50{}^\circ =\tan 40{}^\circ +2\tan 10{}^\circ .$ (iii) $\displaystyle \tan 70{}^\circ =2\tan 50{}^\circ +\tan 20{}^\circ .$ (iv) $\displaystyle \tan 3A-\tan 2A-\tan A=\tan 3A\tan 2A\tan A.$
Show/Hide Solution
 (i) $\displaystyle \ \ \ \ \frac{{\cos 17{}^\circ +\sin 17{}^\circ }}{{\cos 17{}^\circ -\sin 17{}^\circ }}$ $\displaystyle \ \ \ \ \ \ =\frac{{\displaystyle \frac{{\cos 17{}^\circ }}{{\cos 17{}^\circ }}+\displaystyle \frac{{\sin 17{}^\circ }}{{\cos 17{}^\circ }}}}{{\displaystyle \frac{{\cos 17{}^\circ }}{{\cos 17{}^\circ }}-\displaystyle \frac{{\sin 17{}^\circ }}{{\cos 17{}^\circ }}}}$ $\displaystyle \ \ \ \ \ \ =\frac{{1+\tan 17{}^\circ }}{{1+\tan 17{}^\circ }}$ $\displaystyle \ \ \ \ \ \ =\frac{{\tan 45{}^\circ +\tan 17{}^\circ }}{{1+\tan 45{}^\circ \tan 17{}^\circ }}$ $\displaystyle \ \ \ \ \ \ =\tan (45{}^\circ +17{}^\circ )$ $\displaystyle \ \ \ \ \ \ =\tan 62{}^\circ$ (ii) $\displaystyle \ \ \ \ \ \ \tan 50{}^\circ =\tan (40{}^\circ +10{}^\circ )$ $\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \tan 50{}^\circ =\frac{{\tan 40{}^\circ +\tan 10{}^\circ }}{{1-\tan 40{}^\circ \tan 10{}^\circ }}$ $\displaystyle \begin{array}{l}\ \ \ \ \ \ \ \ \ \ \ \tan 50{}^\circ -\tan 50{}^\circ \tan 40{}^\circ \tan 10{}^\circ =\tan 40{}^\circ +\tan 10{}^\circ \\\\\ \ \ \ \ \ \ \ \ \ \ \tan 50{}^\circ =\tan 40{}^\circ +\tan 10{}^\circ +\tan 50{}^\circ \tan 40{}^\circ \tan 10{}^\circ \\\\\ \ \ \ \ \ \ \ \ \ \ \tan 50{}^\circ =\tan 40{}^\circ +\tan 10{}^\circ +\tan (90{}^\circ -40{}^\circ )\tan 40{}^\circ \tan 10{}^\circ \\\\\ \ \ \ \ \ \ \ \ \ \ \tan 50{}^\circ =\tan 40{}^\circ +\tan 10{}^\circ +\cot 40{}^\circ \tan 40{}^\circ \tan 10{}^\circ \end{array}$ $\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \tan 50{}^\circ =\tan 40{}^\circ +\tan 10{}^\circ +\frac{1}{{\tan 40{}^\circ }}(\tan 40{}^\circ \tan 10{}^\circ )$ $\displaystyle \therefore \ \ \ \ \ \ \ \ \tan 50{}^\circ =\tan 40{}^\circ +2\tan 10{}^\circ$ (iii) $\displaystyle \ \ \ \ \tan 70{}^\circ =\tan (50{}^\circ +20{}^\circ )$ $\displaystyle \ \ \ \ \ \ \ \ \ \ \tan 70{}^\circ =\frac{{\tan 50{}^\circ +\tan 20{}^\circ }}{{1-\tan 50{}^\circ \tan 20{}^\circ }}$ $\displaystyle \ \ \ \ \ \ \ \ \ \ \tan 70{}^\circ -\tan 70{}^\circ \tan 50{}^\circ \tan 20{}^\circ =\tan 50{}^\circ +\tan 20{}^\circ$ $\displaystyle \ \ \ \ \ \ \ \ \ \ \tan 70{}^\circ =\tan 50{}^\circ +\tan 20{}^\circ +\tan 70{}^\circ \tan 50{}^\circ \tan 20{}^\circ$ $\displaystyle \ \ \ \ \ \ \ \ \ \ \tan 70{}^\circ =\tan 50{}^\circ +\tan 20{}^\circ +\tan (90{}^\circ -20{}^\circ )\tan 50{}^\circ \tan 20{}^\circ$ $\displaystyle \ \ \ \ \ \ \ \ \ \ \tan 70{}^\circ =\tan 50{}^\circ +\tan 50{}^\circ +\cot 20{}^\circ \tan 50{}^\circ \tan 20{}^\circ$ $\displaystyle \ \ \ \ \ \ \ \ \ \ \tan 70{}^\circ =\tan 50{}^\circ +\tan 20{}^\circ +\frac{1}{{\tan 20{}^\circ }}(\tan 50{}^\circ \tan 20{}^\circ )$ $\displaystyle \therefore \ \ \ \ \ \ \tan 70{}^\circ =2\tan 50{}^\circ +\tan 20{}^\circ$ (iv) $\displaystyle \ \ \tan 3A=\tan (2A+A)$ $\displaystyle \ \ \ \ \ \ \ \ \tan 3A=\frac{{\tan 2A+\tan A}}{{1-\tan 2A\tan A}}$ $\displaystyle \ \ \ \ \ \ \ \ \tan 3A-\tan 3A\tan 2A\tan A=\tan 2A+\tan A$ $\displaystyle \therefore \ \ \ \ \tan 3A-\tan 2A-\tan A=\tan 3A\tan 2A\tan A$

 8.        (i) If $\displaystyle A+B=45{}^\circ,$ prove that $\displaystyle (1+\tan A)(1+\tan B)=2.$ (ii) If $\displaystyle A+B=45{}^\circ,$ prove that $\displaystyle (\cot A - 1)(\cot B - 1)=2.$ (iii) If $\displaystyle A-B=45{}^\circ,$ prove that $\displaystyle (1+\tan A)(1+\tan B)=2\tan A.$
Show/Hide Solution
 (i) $\displaystyle \ A+B=45{}^\circ$ $\displaystyle \therefore \ \ \ \tan (A+B)=\tan 45{}^\circ =1$ $\displaystyle \therefore \ \ \ \frac{{\tan A+\tan B}}{{1-\tan A\tan B}}=1$ $\displaystyle \begin{array}{l}\therefore \ \ \ \tan A+\tan B=1-\tan A\tan B\\\\\therefore \ \ \ \tan A+\tan B+\tan A\tan B=1\\\\\therefore \ \ \ 1+\tan A+\tan B+\tan A\tan B=2\\\\\therefore \ \ \ \left( {1+\tan A} \right)+\tan B\left( {1+\tan A} \right)=2\\\\\therefore \ \ \ \left( {1+\tan A} \right)(1+\tan B)=2\end{array}$ (ii) $\displaystyle \ A+B=45{}^\circ$ $\displaystyle \therefore \ \ \ \tan (A+B)=\tan 45{}^\circ =1$ $\displaystyle \therefore \ \ \ \frac{{\tan A+\tan B}}{{1-\tan A\tan B}}=1$ $\displaystyle \therefore \ \ \ \tan A+\tan B=1-\tan A\tan B$ $\displaystyle \therefore \ \ \ \frac{1}{{\cot A}}+\frac{1}{{\cot B}}=1-\frac{1}{{\cot A\cot B}}$ $\displaystyle \begin{array}{l}\ \ \ \ \ \text{Multiplying}\ \text{both}\ \text{sides}\ \text{with }\cot A\cot B,\\\\\ \ \ \ \ \cot B+\cot A=\cot A\cot B-1\\\\\ \ \ \ \ \cot A\cot B-\cot B-\cot A=1\\\\\ \ \ \ \ \cot A\cot B-\cot B-\cot A+1=2\\\\\therefore \ \ \ \cot B\left( {\cot A-1} \right)-\left( {\cot A-1} \right)=2\end{array}$ (iii)$\displaystyle \ A-B=45{}^\circ$ $\displaystyle \therefore \ \ \ \tan (A-B)=\tan 45{}^\circ =1$ $\displaystyle \therefore \ \ \ \frac{{\tan A-\tan B}}{{1+\tan A\tan B}}=1$ $\displaystyle \begin{array}{l}\therefore \ \ \ \tan A-\tan B=1+\tan A\tan B\\\\\therefore \ \ \ 1+\tan A\tan B+\tan B=\tan A\\\\\ \ \ \ \ \text{Adding }\tan A\ \text{to both}\ \text{sides,}\ \\\\\ \ \ \ \ 1+\tan A+\tan A\tan B+\tan B=2\tan A\\\\\ \ \ \ \ \left( {1+\tan A} \right)+\tan B\left( {1+\tan A} \right)=2\tan A\end{array}$

 9.        Prove that $\displaystyle \frac{{\tan (A+B)}}{{\cot (A-B)}}=\frac{{{{{\sin }}^{2}}A-{{{\sin }}^{2}}B}}{{{{{\cos }}^{2}}A-{{{\sin }}^{2}}B}}.$
Show/Hide Solution
 $\displaystyle \ \ \ \ \ \ \ \ \ \ \tan (A+B)=\frac{{\sin (A+B)}}{{\cos (A+B)}}$ $\displaystyle \ \ \ \ \ \ \ \ \ \ \tan (A+B)=\frac{{\sin A\cos B+\cos A\sin B}}{{\cos A\cos B-\sin A\sin B}}--(1)$ $\displaystyle \ \ \ \ \ \ \ \ \ \ \text{Similarly,}$ $\displaystyle \ \ \ \ \ \ \ \ \ \ \tan (A-B)=\frac{{\sin A\cos B-\cos A\sin B}}{{\cos A\cos B+\sin A\sin B}}--(2)$ $\displaystyle \begin{array}{l}\ \ \ \ \ \ \ \ \ \ \text{By (1)}\times \text{(2), we get}\\\\\ \ \ \ \ \ \ \ \ \ \tan (A+B)\tan (A-B)\end{array}$ $\displaystyle \ \ \ =\ \ \frac{{{{{\sin }}^{2}}A{{{\cos }}^{2}}B-{{{\cos }}^{2}}A{{{\sin }}^{2}}B}}{{{{{\cos }}^{2}}A{{{\cos }}^{2}}B-{{{\sin }}^{2}}A{{{\sin }}^{2}}B}}$ $\displaystyle \ \ \ =\ \ \frac{{{{{\sin }}^{2}}A(1-{{{\sin }}^{2}}B)-(1-{{{\sin }}^{2}}A){{{\sin }}^{2}}B}}{{{{{\cos }}^{2}}A(1-{{{\sin }}^{2}}B)-(1-{{{\cos }}^{2}}A){{{\sin }}^{2}}B}}$ $\displaystyle \ \ \ =\ \ \frac{{{{{\sin }}^{2}}A-{{{\sin }}^{2}}A{{{\sin }}^{2}}B-{{{\sin }}^{2}}B+{{{\sin }}^{2}}A{{{\sin }}^{2}}B}}{{{{{\cos }}^{2}}A-{{{\cos }}^{2}}A{{{\sin }}^{2}}B-{{{\sin }}^{2}}B+{{{\cos }}^{2}}A{{{\sin }}^{2}}B}}$ $\displaystyle \ \ \ =\ \ \frac{{{{{\sin }}^{2}}A-{{{\sin }}^{2}}B}}{{{{{\cos }}^{2}}A-{{{\sin }}^{2}}B}}$ $\displaystyle \ \ \ \ \ \ \ \ \ \ \frac{{\tan (A+B)}}{{\cot (A-B)}}=\tan (A+B)\tan (A-B)$ $\displaystyle \therefore \ \ \ \ \ \ \frac{{\tan (A+B)}}{{\cot (A-B)}}=\frac{{{{{\sin }}^{2}}A-{{{\sin }}^{2}}B}}{{{{{\cos }}^{2}}A-{{{\sin }}^{2}}B}}$

 10.        If $\displaystyle 3\tan A\tan B=1,$ prove that $\displaystyle 2\cos (A+B)=\cos (A-B).$
Show/Hide Solution
 $\displaystyle \ \ \ \ \ \ \ \ 3\tan A\tan B=1$ $\displaystyle \ \ \ \ \ \ \ \tan A\tan B=\frac{1}{3}$ $\displaystyle \ \ \ \ \ \ \ \cot A\cot B=3$ $\displaystyle \ \ \ \ \ \ \ \frac{{\cos A\cos B}}{{\sin A\sin B}}=3$ $\displaystyle \ \ \ \ \ \ \ \text{By Componendo - Dividendo,}$ $\displaystyle \ \ \ \ \ \ \ \frac{{\cos A\cos B+\sin A\sin B}}{{\cos A\cos B-\sin A\sin B}}=\frac{{3+1}}{{3-1}}$ $\displaystyle \ \ \ \ \ \ \ \frac{{\cos (A-B)}}{{\cos (A+B)}}=2$ $\displaystyle \therefore \ \ \ \ 2\cos (A+B)=\cos (A-B)$

 11.        Prove that $\displaystyle \frac{{\cos 2\theta }}{{\sin \theta }}+\frac{{\sin 2\theta }}{{\cos \theta }}=\operatorname{cosec}\theta$
Show/Hide Solution
 Solution $\displaystyle \begin{array}{l}\ \ \ \ \ \ \ \ \displaystyle \frac{{\cos 2\theta }}{{\sin \theta }}+\displaystyle \frac{{\sin 2\theta }}{{\cos \theta }}\\\\=\ \ \ \displaystyle \frac{{\cos 2\theta \cos \theta }}{{\sin \theta \cos \theta }}+\displaystyle \frac{{\sin 2\theta \sin \theta }}{{\sin \theta \cos \theta }}\\\\=\ \ \ \displaystyle \frac{{\cos 2\theta \cos \theta +\sin 2\theta \sin \theta }}{{\sin \theta \cos \theta }}\\\\=\ \ \ \displaystyle \frac{{\cos 2\theta \cos \theta +\sin 2\theta \sin \theta }}{{\sin \theta \cos \theta }}\\\\=\ \ \ \displaystyle \frac{{\cos \left( {2\theta -\theta } \right)}}{{\sin \theta \cos \theta }}\\\\=\ \ \ \displaystyle \frac{{\cos \theta }}{{\sin \theta \cos \theta }}\\\\=\ \ \ \displaystyle \frac{1}{{\sin \theta }}\\\\=\ \ \ \operatorname{cosec}\theta \end{array}$