Question

सिद्ध कीजिए कि ${\displaystyle \frac{1+\sec \theta -\tan \theta }{1+\sec \theta +\tan \theta }=\frac{1-\sin \theta }{\cos \theta }}$ है।

Answer 1

L.H.S = ${\displaystyle \frac{1+\sec \theta -\tan \theta }{1+\sec \theta +\tan \theta }}$

= ${\displaystyle \frac{1+1/\cos \theta -\sin \theta /\cos \theta }{1+1/\cos \theta +\sin \theta /\cos \theta }}$  ...${\displaystyle [\because \sec \theta =\frac{1}{\cos \theta }\text{and}\tan \theta =\frac{\sin \theta }{\cos \theta }]}$

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

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

= ${\displaystyle \frac{2{\cos }^2 \frac{\theta }{2}-2\sin \frac{\theta }{2}\cdot \cos \frac{\theta }{2}}{2{\cos }^2 \frac{\theta }{2}+2\sin \frac{\theta }{2}\cdot \cos \frac{\theta }{2}}}$  ...${\displaystyle [\because 1+\cos \theta =2{\cos }^2 \frac{\theta }{2}\text{and}\sin \theta =2\sin \frac{\theta }{2}\cos \frac{\theta }{2}]}$

= ${\displaystyle \frac{2{\cos }^2 \frac{\theta }{2}-2\sin \frac{\theta }{2}\cdot \cos \frac{\theta }{2}}{2{\cos }^2 \frac{\theta }{2}+2\sin \frac{\theta }{2}\cdot \cos \frac{\theta }{2}}}$

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

= ${\displaystyle \frac{\cos \frac{\theta }{2}-\sin \frac{\theta }{2}}{\cos \frac{\theta }{2}+\sin \frac{\theta }{2}}\times \frac{(\cos \frac{\theta }{2}-\sin \frac{\theta }{2})}{(\cos \frac{\theta }{2}-\sin \frac{\theta }{2})}}$  ...[युक्तिकरण द्वारा]

= ${\displaystyle \frac{{(\cos \frac{\theta }{2}-\sin \frac{\theta }{2})}^2}{({\cos }^2 \frac{\theta }{2}-{\sin }^2 \frac{\theta }{2})}}$  ...[∵ (a – b)² = a² + b² – 2ab and (a – b)(a + b) = (a² – b²)]

= ${\displaystyle \frac{({\cos }^2 \frac{\theta }{2}+{\sin }^2 \frac{\theta }{2})-(2\sin \frac{\theta }{2}\cdot \cos \frac{\theta }{2})}{\cos \theta }}$   ...${\displaystyle [\because {\cos }^2 \frac{\theta }{2}-{\sin }^2 \frac{\theta }{2}=\cos \theta ]}$

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

= R.H.S

अत: सिद्ध हुआ।