Question

Prove that:

(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)

Answer 1

$$\begin{gathered} \begin{array}{l}\left(\mathrm{i}\right)\mathrm{LH}S=\sin \mathrm{\theta }\cos ({90}^0-\mathrm{\theta })+\sin ({90}^0-\mathrm{\theta })\cos \mathrm{\theta }\\ \end{array} \\ \begin{array}{l}=\sin \mathrm{\theta }\sin \mathrm{\theta }+\cos \mathrm{\theta }\cos \mathrm{\theta }\\ \end{array} \\ \begin{array}{l}={\sin }^2\mathrm{\theta }+{\cos }^2\mathrm{\theta }\\ \end{array} \\ \begin{array}{l}\mathrm{}=1\\ \end{array} \\ \begin{array}{l}=\mathrm{RHS}\\ \end{array} \\ \begin{array}{l}H\mathrm{e}\mathrm{n}\mathrm{c}\mathrm{e}\mathrm{pr}\mathrm{o}\mathrm{v}\mathrm{e}\mathrm{d}.\\ \end{array} \\ \begin{array}{l}\left(\mathrm{ii}\right)\mathrm{LHS=}\frac{\sin \mathrm{\theta }}{\cos ({90}^0-\mathrm{\theta })}+\frac{\cos \mathrm{\theta }}{\sin ({90}^0-\mathrm{\theta })}\\ \end{array} \\ \begin{array}{l}=\frac{\sin \mathrm{\theta }}{\sin \mathrm{\theta }}+\frac{\cos \mathrm{\theta }}{\cos \mathrm{\theta }}\\ \end{array} \\ \begin{array}{l}=1+1\\ \end{array} \\ \begin{array}{l}=2\\ \end{array} \\ \begin{array}{l}=\mathrm{RHS}\\ \end{array} \\ \begin{array}{l}\mathrm{Hence}\mathrm{pr}\mathrm{o}\mathrm{v}\mathrm{e}\mathrm{d}.\\ \mathrm{}\end{array} \\ \begin{array}{l}\mathrm{}\left(\mathrm{iii}\right)\mathrm{LHS}=\frac{\sin \mathrm{\theta }\cos ({90}^0-\mathrm{\theta })\cos \mathrm{\theta }}{\sin ({90}^0-\mathrm{\theta })}+\frac{\cos \mathrm{\theta }\sin ({90}^0-\mathrm{\theta })\sin \mathrm{\theta }}{\cos ({90}^0-\mathrm{\theta })}\\ \end{array} \\ \begin{array}{l}=\frac{\sin \mathrm{\theta }\sin \mathrm{\theta }\cos \mathrm{\theta }}{\cos \mathrm{\theta }}+\frac{\cos \mathrm{\theta }\cos \mathrm{\theta }\sin \mathrm{\theta }}{\sin \mathrm{\theta }}\\ \end{array} \\ \begin{array}{l}={\sin }^2\mathrm{\theta }+{\cos }^2\mathrm{\theta }\\ \end{array} \\ \begin{array}{l}=1\\ \end{array} \\ \begin{array}{l}\begin{array}{l}=\mathrm{RHS}\\ \end{array} \\ \begin{array}{l}H\mathrm{e}\mathrm{n}\mathrm{c}\mathrm{e}\mathrm{pr}\mathrm{o}\mathrm{v}\mathrm{e}\mathrm{d}.\\ \end{array}\\ \end{array} \\ \begin{array}{l}\left(\mathrm{iv}\right)\mathrm{LHS}=\frac{\cos ({90}^0-\mathrm{\theta })\sec ({90}^0-\mathrm{\theta })\tan \mathrm{\theta }}{\cos \mathrm{ec}({90}^0-\mathrm{\theta })\sin ({90}^0-\mathrm{\theta })\cot ({90}^0-\mathrm{\theta })}+\frac{\tan ({90}^0-\mathrm{\theta })}{\cot \mathrm{\theta }}\\ \end{array} \\ \begin{array}{l}=\frac{\sin \mathrm{\theta }\cos \mathrm{ec}\mathrm{\theta }\tan \mathrm{\theta }}{\sec \mathrm{\theta }\cos \mathrm{\theta }\tan \mathrm{\theta }}+\frac{\cot \mathrm{\theta }}{\cot \mathrm{\theta }}\\ \end{array} \\ \begin{array}{l}\mathrm{}=1+1\\ \end{array} \\ \begin{array}{l}=2\\ \end{array} \\ \begin{array}{l}=\mathrm{RHS}\\ \end{array} \\ \begin{array}{l}H\mathrm{e}\mathrm{n}\mathrm{c}\mathrm{e}\mathrm{pr}\mathrm{o}\mathrm{v}\mathrm{e}\mathrm{d}.\\ \end{array} \\ \begin{array}{l}\left(\mathrm{v}\right)\mathrm{LHS}=\frac{\cos ({90}^0-\mathrm{\theta })}{1+\sin ({90}^0-\mathrm{\theta })}+\frac{1+\sin ({90}^0-\mathrm{\theta })}{\cos ({90}^0-\mathrm{\theta })}\\ \end{array} \\ \begin{array}{l}\mathrm{}=\frac{\sin \mathrm{\theta }}{1+\cos \mathrm{\theta }}+\frac{1+\cos \mathrm{\theta }}{\sin \mathrm{\theta }}\\ \end{array} \\ \begin{array}{l}=\frac{{\sin }^2\mathrm{\theta }+{(1+\cos \mathrm{\theta })}^2}{(1+\cos \mathrm{\theta })\sin \mathrm{\theta }}\\ \end{array} \\ \begin{array}{l}=\frac{{\sin }^2\mathrm{\theta }+1+{\cos }^2\mathrm{\theta }+2\cos \mathrm{\theta }}{(1+\cos \mathrm{\theta })\sin \mathrm{\theta }}\\ \end{array} \\ \begin{array}{l}=\frac{1+1+2\cos \mathrm{\theta }}{(1+\cos \mathrm{\theta })\sin \mathrm{\theta }}\\ \end{array} \\ \begin{array}{l}=\frac{2+2\cos \mathrm{\theta }}{(1+\cos \mathrm{\theta })\sin \mathrm{\theta }}\\ \end{array} \\ \begin{array}{l}=\frac{2(1+\cos \mathrm{\theta })}{(1+\cos \mathrm{\theta })\sin \mathrm{\theta }}\\ \end{array} \\ \begin{array}{l}=2\frac{1}{\sin \mathrm{\theta }}\\ \end{array} \\ \begin{array}{l}=2\cos \mathrm{ec}\mathrm{\theta }\\ \end{array} \\ \begin{array}{l}=\mathrm{RHS}\\ \end{array} \\ \begin{array}{l}H\mathrm{e}\mathrm{n}\mathrm{c}\mathrm{e}\mathrm{pr}\mathrm{o}\mathrm{v}\mathrm{e}\mathrm{d}.\\ \end{array} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{vi}\right)\mathrm{LHS}=\frac{\sec \left(90°-\theta \right)\mathrm{cosec}\theta -\tan \left(90°-\theta \right)\cot \theta +{\cos }^{2}25°+{\cos }^{2}65°}{3\tan 27°\tan 63°} \\ =\frac{\mathrm{cosec}\theta \mathrm{cosec}\theta -\cot \theta \cot \theta +{\sin }^2\left(90°-25°\right)+{\cos }^{2}65°}{3\tan 27°\cot \left(90°-63°\right)} \\ =\frac{{\mathrm{cosec}}^2\theta -{\cot }^2\theta +{\sin }^{2}65°+{\cos }^{2}65°}{3\tan 27°\cot 27°} \\ =\frac{1+1}{3\times \tan 27°\times {\displaystyle \frac{1}{\tan 27°}}} \\ =\frac{2}{3} \\ =\mathrm{RHS} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{vii}\right)\mathrm{LHS}=\cot \theta \tan \left(90°-\theta \right)-\sec \left(90°-\theta \right)\mathrm{cosec}\theta +\sqrt{3}\tan 12°\tan 60°\tan 78° \\ =\cot \theta \cot \theta -\mathrm{cosec}\theta \mathrm{cosec}\theta +\sqrt{3}\tan 12°\times \sqrt{3}\times \cot \left(90°-78°\right) \\ ={\cot }^2\theta -{\mathrm{cosec}}^2\theta +3\tan 12°\cot 12° \\ =-1+3\times \tan 12°\times \frac{1}{\tan 12°} \\ =-1+3 \\ =2 \\ =\mathrm{RHS} \end{gathered}$$