TBQ - Prove the following VI — task. Maths TNSB Mentoring, Class 10 Crash course.

Prove that \(cot^2 \ P - cot^2 \ Q = \frac{cos^2 \ P - cos^2 \ Q}{sin^2 \ P \ sin^2 \ Q}\).

Proof:

Answer variants:

\(1 - sec^2 P\)

\(cos^2 P - cos^2 Q\)

\(1 - cos^2 P\)

\(1 - cosec^2 P\)

\(1 - cos^2 Q\)

\begin{array}{l} LHS = \cot^{2} P - \cot^{2} Q \\ = \frac{\cos^{2} P}{\sin^{2} P} - \frac{\cos^{2} Q}{\sin^{2} Q} \\ = \frac{\cos^{2} P \sin^{2} Q - \cos^{2} Q \sin^{2} P}{\sin^{2} P \sin^{2} Q} \\ = \frac{\cos^{2} P (i) - \cos^{2} Q (i)}{\sin^{2} P \sin^{2} Q} \\ = \frac{\cos^{2} P - \cos^{2} P \cos^{2} Q - \cos^{2} Q + \cos^{2} Q \cos^{2} P}{\sin^{2} P \sin^{2} Q} \\ = \frac{i}{\sin^{2} P \sin^{2} Q} \end{array}

Did you find an error?

10. TBQ - Prove the following VI