If-cos-1-sin-a-a-pi-2-2kpi-So-tan-2-A-a-a-1-D-a-1-a-1-B-1-a-1-E-a-a-1-C-a-1-a

Question Number 10955 by Joel576 last updated on 04/Mar/17

If \frac{\cos θ}{1 - \sin θ} = a a \neq \frac{π}{2} + 2 k π

So, \tan \frac{θ}{2} = \dots

(A) \frac{a}{a + 1} (D) \frac{a + 1}{a - 1}

(B) \frac{1}{a + 1} (E) \frac{a}{a - 1}

(C) \frac{a - 1}{a + 1}

Answered by ajfour last updated on 04/Mar/17

\frac{\sin (\frac{π}{2} - θ)}{1 - \cos (\frac{π}{2} - θ)} = \frac{2 \sin (\frac{π}{4} - \frac{θ}{2}) \cos (\frac{π}{4} - \frac{θ}{2})}{2 \sin^{2} (\frac{π}{4} - \frac{θ}{2})}

= \cot (\frac{π}{4} - \frac{θ}{2}) = a

\tan (\frac{π}{4} - \frac{θ}{2}) = \frac{1}{a}; l e t \tan (\frac{θ}{2}) = p

\frac{1 - p}{1 + p} = \frac{1}{a} \Rightarrow a - a p = 1 + p

p = \frac{a - 1}{a + 1} .

Commented by Joel576 last updated on 05/Mar/17

t h a n k y o u v e r y m u c h

Answered by ridwan balatif last updated on 04/Mar/17

another way

{(\frac{\cos θ}{1 - \sin θ})}^{2} = {(a)}^{2}

\frac{\cos^{2} θ}{{(1 - \sin θ)}^{2}} = a^{2}

\frac{(1 - \sin^{2} θ)}{{(1 - \sin θ)}^{2}} = a^{2}

\frac{(1 - \sin θ) (1 + \sin θ)}{(1 - \sin θ) (1 - \sin θ)} = a^{2}

\frac{1 + \sin θ}{1 - \sin θ} = a^{2}

\sin θ = \frac{a^{2} - 1}{a^{2} + 1}

\cos θ = \frac{2 a}{a^{2} + 1}

remember : \tan \frac{θ}{2} = \frac{\sin θ}{1 + \cos θ}

\tan \frac{θ}{2} = \frac{\frac{a^{2} - 1}{a^{2} + 1}}{1 + \frac{2 a}{a^{2} + 1}}

\tan \frac{θ}{2} = \frac{\frac{a^{2} - 1}{a^{2} + 1}}{\frac{a^{2} + 2 a + 1}{a^{2} + 1}}

\tan \frac{θ}{2} = \frac{a^{2} - 1}{a^{2} + 2 a + 1}

\tan \frac{θ}{2} = \frac{(a - 1) (a + 1)}{{(a + 1)}^{2}}

\tan \frac{θ}{2} = \frac{a - 1}{a + 1}

Answer : (C)

Commented by Joel576 last updated on 05/Mar/17

t h a n k y o u v e r y m u c h

Facebook Tweet Pin