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Trigonometric Functions

Multiple Choice Questions

141.

If $α and β$ are the solutions of $atan (θ) + bsec (θ) = c$ , then $\tan (α + β)$ is equal

$\frac{2 ac}{a^{2} - c^{2}}$
$\frac{2 ac}{c^{2} - a^{2}}$
$\frac{2 ac}{a^{2} + c^{2}}$
None of these

142.

If $\sin^{- 1} (\frac{x}{5}) + \csc^{- 1} (\frac{5}{4}) = \frac{π}{2}$ , then x is equal to

143.
If $\frac{\sin (A)}{\sin (C)} = \frac{\sin (A - B)}{\sin (B - C)}$ , then a², b², c² are in

AP

GP

HP

None of these

$\frac{\sin (A)}{\sin (C)} = \frac{\sin (A - B)}{\sin (B - C)} \frac{\sin (A)}{\sin (C)} = \frac{\sin (A) . \cos (B) - \cos (A) . \sin (B)}{\sin (B) . \cos (C) - \cos (B) . \sin (C)} \frac{aK}{cK} = \frac{aKcos (B) - bKcos (A)}{bKcos (C) - cKcos (B)} \{∵ \frac{\sin (A)}{a} = \frac{\sin (B)}{b} = \frac{\sin (C)}{c} = K\} \frac{a}{c} = \frac{acos (B) - bcos (A)}{bcos (C) - ccos (B)} \{∵ \cos (A) = \frac{b^{2} + c^{2} - a^{2}}{2 bc} \cos (B) = \frac{a^{2} + c^{2} - b^{2}}{2 ac} \cos (C) = \frac{a^{2} + b^{2} - c^{2}}{2 ab}\} \frac{a}{c} = \frac{\frac{(a^{2} + c^{2} - b^{2})}{2 c} - \frac{(b^{2} + c^{2} - a^{2})}{2 c}}{\frac{(a^{2} + b^{2} - c^{2})}{2 a} - \frac{(a^{2} + c^{2} - b^{2})}{2 a}}$

$= \frac{a}{c} [\frac{a^{2} + c^{2} - b^{2} - c^{2} - b^{2} + a^{2}}{a^{2} + b^{2} - c^{2} - a^{2} - c^{2} + b^{2}}] 1 = \frac{2 a^{2} - 2 b^{2}}{2 b^{2} - 2 c^{2}} \Rightarrow 1 = \frac{a^{2} - b^{2}}{b^{2} - c^{2}} \Rightarrow b^{2} - c^{2} = a^{2} - b^{2} \Rightarrow 2 b^{2} = a^{2} + b^{2} Hence, a^{2}, b^{2}, c^{2} are in AP .$