Question

निम्नलिखित सर्वसमिकाएँ को सिद्ध कीजिए, जहाँ वे कोण जिनके लिए व्यंजक परिभाषित किया गया है, न्यून कोण हैं:

${\displaystyle \frac{\tan \theta }{1-\cot \theta }+\frac{\cot \theta }{1-\tan \theta }=1+\sec \theta \cos ec \theta }$

[संकेत: व्यंजक को sin θ और cosθ के पदों में लिखिए]

Answer 1

L.H.S

= ${\displaystyle \frac{\tan \theta }{1-\cot \theta }+\frac{\cot \theta }{1-\tan \theta }}$

= ${\displaystyle \frac{\frac{\sin \theta }{\cos \theta }}{1-\frac{\cos \theta }{\sin \theta }}+\frac{\frac{\cos \theta }{\sin \theta }}{1-\frac{\sin \theta }{\cos \theta }}}$

= ${\displaystyle \frac{\frac{\sin \theta }{\cos \theta }}{\frac{\sin \theta -\cos \theta }{\sin \theta }}+\frac{\frac{\cos \theta }{\sin \theta }}{\frac{\cos \theta -\sin \theta }{\cos \theta }}}$

= ${\displaystyle \frac{{\sin }^2\theta }{\cos \theta (\sin \theta -\cos \theta )}-\frac{{\cos }^2\theta }{\sin \theta (\sin \theta -\cos \theta )}}$

= ${\displaystyle \frac{1}{\sin \theta -\cos \theta }[\frac{{\sin }^2\theta }{\cos \theta }-\frac{{\cos }^2\theta }{\sin \theta }]}$

= ${\displaystyle \left(\frac{1}{\sin \theta -\cos \theta }\right)\left[\frac{{\sin }^3\theta -{\cos }^3\theta }{\sin \theta \cos \theta }\right]}$

= ${\displaystyle \left(\frac{1}{\sin \theta -\cos \theta }\right)\left[\frac{(\sin \theta -\cos \theta )({\sin }^2\theta +{\cos }^2\theta +\sin \theta \cos \theta )}{\sin \theta \cos \theta }\right]}$

= ${\displaystyle \frac{(1+\sin \theta \cos \theta )}{\left(\sin \theta \cos \theta \right)}}$

= sec θ cosec θ + 1

= R.H.S