TBQ - Trigonometric identities — task. Maths TNSB Mentoring, Class 10 Crash course.

If \(\frac{cot \ \alpha}{cot \ \beta}\) \(= p\) and \(\frac{cot \ \alpha}{cosec \ \beta}\) \(= q\), then prove that

q^{2} = (p^{2} - q^{2}) \cot^{2} β

Proof:

Consider \(\frac{cot \ \alpha}{cot \ \beta}\) \(= p\)

---- (\(1\))

Consider \(\frac{cos \ \alpha}{sin \ \beta} = n\)

---- (\(2\))

Using equation (\(2\)) in equation (\(1\)), we get:

Squaring on both sides, we have:

Hence, we proved.

Answer variants:

\cot α = q cosec β

q^{2} + q^{2} \cot^{2} β = p^{2} \cot^{2} β

q^{2} (1 + \cot^{2} β) = p^{2} \cot^{2} β

q^{2} {cosec}^{2} β = p^{2} \cot^{2} β

\cot α = p \cot β

q^{2} = (p^{2} - q^{2}) \cot^{2} β

q^{2} = p^{2} \cot^{2} β - q^{2} \cot^{2} β

q cosec β = p \cot β

Did you find an error?

8. TBQ - Trigonometric identities