25 (a) Prove that sqrt((1+ A)/(1- A)) = A + A

Alright Rajnish — let's do this. Here's the solution for question 25 (a) and (b): 25 (a) Prove that $\sqrt{\frac{1+\sin A}{1-\sin A}} = \sec A + \tan A$ Step 1: Start with the Left Hand Side (LHS). $$ \text{LHS} = \sqrt{\frac{1+\sin A}{1-\sin A}} $$ Step 2: Multiply the numerator and denominator inside the square root by the conjugate of the denominator, $(1+\sin A)$. $$ \text{LHS} = \sqrt{\frac{1+\sin A}{1-\sin A} \times \frac{1+\sin A}{1+\sin A}} $$ $$ \text{LHS} = \sqrt{\frac{(1+\sin A)^2}{1^2 - \sin^2 A}} $$ Step 3: Use the identity $1 - \sin^2 A = \cos^2 A$. $$ \text{LHS} = \sqrt{\frac{(1+\sin A)^2}{\cos^2 A}} $$ Step 4: Take the square root of the numerator and the denominator. $$ \text{LHS} = \frac{\sqrt{(1+\sin A)^2}}{\sqrt{\cos^2 A}} $$ $$ \text{LHS} = \frac{1+\sin A}{\cos A} $$ Step 5: Separate the terms. $$ \text{LHS} = \frac{1}{\cos A} + \frac{\sin A}{\cos A} $$ Step 6: Express in terms of $\sec A$ and $\tan A$. $$ \text{LHS} = \sec A + \tan A $$ This is equal to the Right Hand Side (RHS). $$ \text{LHS} = \text{RHS} $$ Hence, proved. $\boxed{\text{Proved}}$ 25 (b) Evaluate $\frac{3 \cos^2 30^\circ - 6 \operatorname{cosec}^2 30^\circ}{\tan^2 60^\circ}$ Step 1: Recall the values of the trigonometric functions. $$ \cos 30^\circ = \frac{\sqrt{3}}{2} $$ $$ \operatorname{cosec} 30^\circ = \frac{1}{\sin 30^\circ} = \frac{1}{1/2} = 2 $$ $$ \tan 60^\circ = \sqrt{3} $$ Step 2: Substitute these values into the expression. $$ \frac{3 \left(\frac{\sqrt{3}}{2}\right)^2 - 6 (2)^2}{(\sqrt{3})^2} $$ Step 3: Calculate the squares. $$ \frac{3 \left(\frac{3}{4}\right) - 6 (4)}{3} $$ $$ \frac{\frac{9}{4} - 24}{3} $$ Step 4: Simplify the numerator. $$ \frac{\frac{9}{4} - \frac{96}{4}}{3} $$ $$ \frac{-\frac{87}{4}}{3} $$ Step 5: Perform the division. $$ -\frac{87}{4} \times \frac{1}{3} $$ $$ -\frac{87}{12} $$ Step 6: Simplify the fraction by dividing the numerator and denominator by 3. $$ -\frac{29}{4} $$ The evaluated value is $\boxed{-\frac{29}{4}}$. Send me the next one 📸