Q4(ix):

Prove the following identities, where the angles involved are acute angles for which the expressions are defined. (ix) $(\text{cosec } A – \sin A) (\sec A – \cos A) = \frac{1}{\tan A + \cot A}$ [Hint : Simplify LHS and RHS separately]

Given: The trigonometric identity $(\text{cosec } A - \sin A)(\sec A - \cos A) = \frac{1}{\tan A + \cot A}$, where $A$ is an acute angle.

To Prove: The Left Hand Side (LHS) is equal to the Right Hand Side (RHS).

Step 1: Simplifying the Left Hand Side (LHS)

LHS = $(\text{cosec } A - \sin A)(\sec A - \cos A)$

Using the reciprocal identities $\text{cosec } A = \frac{1}{\sin A}$ and $\sec A = \frac{1}{\cos A}$:

LHS = $\left( \frac{1}{\sin A} - \sin A \right) \left( \frac{1}{\cos A} - \cos A \right)$

Taking the common denominator for each bracket:

LHS = $\left( \frac{1 - \sin^2 A}{\sin A} \right) \left( \frac{1 - \cos^2 A}{\cos A} \right)$

Applying the Pythagorean identity $\sin^2 A + \cos^2 A = 1$, which implies $1 - \sin^2 A = \cos^2 A$ and $1 - \cos^2 A = \sin^2 A$:

LHS = $\left( \frac{\cos^2 A}{\sin A} \right) \left( \frac{\sin^2 A}{\cos A} \right)$

Canceling the common terms in the numerator and denominator:

LHS = $\cos A \cdot \sin A$

Step 2: Simplifying the Right Hand Side (RHS)

RHS = $\frac{1}{\tan A + \cot A}$

Using the quotient identities $\tan A = \frac{\sin A}{\cos A}$ and $\cot A = \frac{\cos A}{\sin A}$:

RHS = $\frac{1}{\frac{\sin A}{\cos A} + \frac{\cos A}{\sin A}}$

Finding the common denominator in the denominator of the fraction:

RHS = $\frac{1}{\frac{\sin^2 A + \cos^2 A}{\cos A \sin A}}$

Applying the Pythagorean identity $\sin^2 A + \cos^2 A = 1$:

RHS = $\frac{1}{\frac{1}{\cos A \sin A}}$

By the property of reciprocals of fractions ($\frac{1}{1/x} = x$):

RHS = $\sin A \cos A$

Step 3: Conclusion

Since LHS = $\sin A \cos A$ and RHS = $\sin A \cos A$, we have shown that LHS = RHS.

Final Answer: Hence, it is proved that $(\text{cosec } A - \sin A)(\sec A - \cos A) = \frac{1}{\tan A + \cot A}$.