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if cosec\theta =\frac{p+q}{p-q}\; \; (p\neq q\neq 0),\; \; then\; \; \left | cot\left ( \frac{\pi }{4}+\frac{\theta }{2} \right ) \right |    is equal to :

  • Option 1)

    \sqrt{\frac{p}{q}}\; \;

  • Option 2)

    \sqrt{\frac{q}{p}}\;

  • Option 3)

    \; \sqrt{pq}

  • Option 4)

    pq

 

Answers (1)

best_answer

As we learnt in 

Trigonometric Ratios of Submultiples of an Angle -

Trigonometric ratios of submultiples of an angle 1

- wherein

This shows the formulae for half angles and their doubles.

 

 cosec \theta =\frac{p+q}{p-q}\Rightarrow sin\theta =\frac{p-q}{p+q}

\cos \theta =\frac{\sqrt{\left ( p+q \right )^{2}-\left ( p-q \right )^{2}}}{p+q}

\cos \theta =\frac{2\sqrt{\left | pq \right |}}{p+q} \: \: \: \: \cdot \cdot \cdot (1)

Now, \left | \cot \left ( \frac{\pi }{4}+\frac{\theta }{2} \right ) \right |= \: \left | \frac{1}{tan\left ( \frac{\pi }{4}+\frac{\theta }{2} \right )} \right |=\left | \frac{1-tan\frac{\theta }{2}}{1+tan\frac{\theta }{2}} \right |

=\left | \frac{\cos \frac{\theta }{2}-\sin\frac{\theta}{2}}{\cos\frac{\theta }{2}+\sin\frac{\theta}{2}}\times \frac{\cos \frac{\theta }{2}-\sin\frac{\theta}{2}}{\cos\frac{\theta }{2}-\sin\frac{\theta}{2}} \right |

= \left | \frac{cos^{2}\frac{\theta }{2}+\sin ^{2}\frac{\theta}{2}-2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}{\cos ^{2}\frac{\theta }{2}-\sin ^{2}\frac{\theta}{2}} \right |= \left | \frac{1-\sin \theta }{\cos \theta } \right | \: \: \: \cdot \cdot (2)

From (1) and (2), \left | cot\left ( \frac{\pi }{4}+\frac{\theta }{2} \right ) \right |= \frac{\frac{2q}{p+q}}{\frac{2\sqrt{pq}}{p+q}}=\sqrt{\frac{q}{p}}


Option 1)

\sqrt{\frac{p}{q}}\; \;

This option is incorrect

Option 2)

\sqrt{\frac{q}{p}}\;

This option is correct

Option 3)

\; \sqrt{pq}

This option is incorrect

Option 4)

pq

This option is incorrect

Posted by

Aadil

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