Question 4

Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

(i) $(\csc \theta - \cot \theta)^2 = \dfrac{1 - \cos \theta}{1 + \cos \theta}$

(ii) $\dfrac{\cos A}{1 - \sin A} = \dfrac{2 \sec A}{1 + \sin A \cos A}$

(iii) $\dfrac{\tan \theta + \cot \theta}{\sec \theta + \csc \theta} = \dfrac{1}{\cot \theta - \theta}$ [Hint: Write the expression in terms of $\sin \theta$ and $\cos \theta$]

(iv) $\dfrac{2}{1 + \sec A} + \sin A = \dfrac{\sec A}{1 - \cos A}$ [Hint: Simplify LHS and RHS separately]

(v) $\dfrac{\cos A - \sin A + 1}{\csc A + \cot A} = \dfrac{\cos A + \sin A - 1}{\csc A - \cot A}$ using the identity $\csc^2 A = 1 + \cot^2 A$.

(vi) $\dfrac{1}{\sin A} + \sec A + \tan A = \dfrac{1}{1 - \sin A}$

(vii) $\dfrac{\sin^2 \theta - \sin \theta \tan^2 \theta}{\cos^2 \theta} = \tan^2 \theta$

(viii) $(\sin A + \csc A)^2 + (\cos A + \sec A)^2 = 7 + \tan^2 A + \cot^2 A$

(ix) $\dfrac{1}{(\csc A - \sin A)(\sec A - \cos A)} = \tan A + \cot A$ [Hint: Simplify LHS and RHS separately]

(x) $\dfrac{1 + \tan A}{1 - \tan A} = \dfrac{1 - \cot A}{1 + \cot A} = \tan^2 A$