J P Mohindru and Bharat Mohindru Solutions for Chapter: Introduction to Trigonometry, Exercise 4: EXERCISE

Prove that $\frac{\tan A+\sin A}{\tan A-\sin A}=\frac{\sec A+1}{\sec A-1}$

Prove that ${\sin }^{2}A{\cos }^{2}B-{\cos }^{2}A{\sin }^{2}B={\sin }^{2}A-{\sin }^{2}B$

Prove that ${\tan }^{2}A{\sec }^{2}B-{\sec }^{2}A{\tan }^{2}B={\tan }^{2}A-{\tan }^{2}B$

Prove that ${\tan }^{2}A{\sec }^{2}B-{\sec }^{2}A{\tan }^{2}B={\tan }^{2}A-{\tan }^{2}B$

Prove that ${\cot }^{2}A{\mathrm{cosec}}^{2}B-{\cot }^{2}B{\mathrm{cosec}}^{2}A={\cot }^{2}A-{\cot }^{2}B$

Prove that $\frac{\sec \theta -1}{\sec \theta +1}={\left(\frac{{\displaystyle \sin \theta }}{{\displaystyle 1+\cos \theta }}\right)}^2$

Prove that ${\left(\frac{\sin \theta +1-\cos \theta }{\cos \theta -1+\sin \theta }\right)}^2=\frac{{\displaystyle 1+\sin \theta }}{{\displaystyle 1-\sin \theta }}$

Prove that ${\left(\frac{1+\cos \theta +\sin \theta }{1+\cos \theta -\sin \theta }\right)}^2=\frac{1+\sin \theta }{1-\sin \theta }$