Question

12,13,14,15all the sub parts

​Q12. Without using trigonometrical tables, evaluate :


(i) $2{\left(\frac{\tan 35°}{\cot 55°}\right)}^2+\left(\frac{\cot 55°}{\tan 35°}\right)-3\left(\frac{\sec 40°}{\mathrm{cosec}50°}\right)$ (ii) $\frac{\sin 35°\cos 55°+\cos 35°\sin 55°}{{\mathrm{cosec}}^{2}10°-{\tan }^{2}80°}$


(iii) sin²34° + sin²56° + 2 tan 18° tan 72° – cot² 30°.


Q13. Prove the following :


(i) $\frac{\cos \mathrm{\theta }}{\sin \left(90°-\mathrm{\theta }\right)}+\frac{\sin {\displaystyle }{\displaystyle \mathrm{\theta }}}{\cos {\displaystyle }{\displaystyle \left(90°-\mathrm{\theta }\right)}}=2$


(ii) cos $\mathrm{\theta }\sin \left(90°-\mathrm{\theta }\right)+\sin \mathrm{\theta }\cos \left(90°-\mathrm{\theta }\right)=1$


(iii) $\frac{\tan \theta }{\tan \left(90°-\theta \right)}+\frac{{\displaystyle \sin \left(90°-\theta \right)}}{{\displaystyle \cos \theta }}={\sec }^2\mathrm{\theta }$.


Q14. Prove the following :


(i) $\frac{\cos \left(90°-\mathrm{A}\right)\sin \left(90°-\mathrm{A}\right)}{\tan \left(90°-\mathrm{A}\right)}=1-{\cos }^2\mathrm{A}$


(ii) $\frac{\sin \left(90°-\mathrm{A}\right)}{\mathrm{cosec}\left(90°-\mathrm{A}\right)}+\frac{\cos {\displaystyle }{\displaystyle \left(90°-\mathrm{A}\right)}}{\sec \left(90°-\mathrm{A}\right)}=1$


Q15. Simplify the following :

(i) $\frac{\cos \mathrm{\theta }}{\sin \left(90°-0\right)}+\frac{\cos {\displaystyle }{\displaystyle \left(90°-\mathrm{\theta }\right)}}{\sec \left(90°-\mathrm{\theta }\right)}-3{\tan }^{2}30°$

Answer 1

Dear student
$$\begin{gathered} \mathrm{Que}15 \\ \frac{\mathrm{cos\theta }}{\sin \left(90°-0\right)}+\frac{\cos \left(90°-\mathrm{\theta }\right)}{\sec \left(90°-\mathrm{\theta }\right)}-3{\tan }^{2}30° \\ =\frac{\mathrm{cos\theta }}{\sin 90°}+\frac{\mathrm{sin\theta }}{\mathrm{cosec\theta }}-3{\left(\frac{1}{\sqrt{3}}\right)}^2 \\ =\mathrm{cos\theta }+\mathrm{sin\theta }\mathrm{sin\theta }-\frac{3}{3} \\ =\mathrm{cos\theta }+{\sin }^2\mathrm{\theta }-1 \\ \mathrm{Properties}\mathrm{used}: \\ 1)\sin (90°-\mathrm{x})=\mathrm{cosx} \\ 2)\cos (90°-\mathrm{x})=\mathrm{sinx} \\ 3)\sec (90°-\mathrm{x})=\mathrm{cosecx} \end{gathered}$$
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