If-secA-tanA-Q-than-prove-that-cosecA-1-Q-2-1-Q-2-

Question Number 173894 by azadsir last updated on 20/Jul/22

If secA - tanA = Q than prove that,

cosecA = \frac{1 + Q^{2}}{1 - Q^{2}}

Commented by cortano1 last updated on 20/Jul/22

Q = \frac{1 - \sin A}{\cos A} \Rightarrow Q^{2} \cos^{2} A = 1 + \sin^{2} A - 2 \sin A

\Rightarrow \sin^{2} A - Q^{2} (1 - \sin^{2} A) - 2 \sin A + 1 = 0

\Rightarrow (1 + Q^{2}) \sin^{2} A - 2 \sin A + 1 - Q^{2} = 0

\Rightarrow \sin A = \frac{2 \pm \sqrt{4 - 4 (1 + Q^{2}) (1 - Q^{2})}}{2 (1 + Q^{2})}

\Rightarrow \sin A = \frac{2 \pm \sqrt{4 - 4 (1 - Q^{4})}}{2 (1 + Q^{2})}

\Rightarrow \csc A = \frac{2 (1 + Q^{2})}{2 \pm 2 Q^{2}} = \frac{1 + Q^{2}}{1 \pm Q^{2}}

\Rightarrow \sin A \neq 1 \Rightarrow \csc A = \frac{1 + Q^{2}}{1 - Q^{2}}

Commented by azadsir last updated on 20/Jul/22

Thank you

Answered by blackmamba last updated on 20/Jul/22

Q = \frac{1 - \sin A}{\cos A} = \frac{\cos \frac{1}{2} A - \sin \frac{1}{2} A}{\cos \frac{1}{2} A + \sin \frac{1}{2} A}

Q = \frac{1 - \tan \frac{1}{2} A}{1 + \tan \frac{1}{2} A}

Q + Q \tan \frac{1}{2} A = 1 - \tan \frac{1}{2} A

\tan \frac{1}{2} A = \frac{1 - Q}{1 + Q}

{\begin{cases} \sin \frac{1}{2} A = \frac{1 - Q}{\sqrt{2 + 2 Q^{2}}} \\ \cos \frac{1}{2} = \frac{1 + Q}{\sqrt{2 + 2 Q^{2}}} \end{cases}

\sin A = \frac{2 (1 - Q^{2})}{2 (1 + Q^{2})}

\frac{1}{\sin A} = \frac{1 + Q^{2}}{1 - Q^{2}}

Commented by azadsir last updated on 20/Jul/22

Thank you

Commented by Tawa11 last updated on 21/Jul/22

Great sirs

Answered by BaliramKumar last updated on 21/Jul/22

s e c A - t a n A = Q

\frac{1 - s i n A}{c o s A} = Q

\frac{{(1 - s i n A)}^{2}}{{c o s}^{2} A} = Q^{2}

\frac{{(1 - s i n A)}^{2}}{1 - {s i n}^{2} A} = Q^{2}

\frac{{(1 - s i n A)}^{2}}{(1 - s i n A) (1 + s i n A)} = Q^{2}

\frac{(1 - s i n A)}{(1 + s i n A)} = Q^{2}

\frac{(1 + s i n A)}{(1 - s i n A)} = \frac{1}{Q^{2}}

\frac{(1 + s i n A) + (1 - s i n A)}{(1 + s i n A) - (1 - s i n A)} = \frac{1 + Q^{2}}{1 - Q^{2}} [∵ I f \frac{a}{b} = \frac{c}{d} t h e n \frac{a + b}{a - b} = \frac{c + d}{c - d}]

\frac{2}{2 s i n A} = \frac{1 + Q^{2}}{1 - Q^{2}}

\frac{1}{s i n A} = \frac{1 + Q^{2}}{1 - Q^{2}}

c o s e c A = \frac{1 + Q^{2}}{1 - Q^{2}}