Q85 / 180 · COMEDK 2023 (Evening Shift) · Mathematics
$\text { If } \operatorname{cosec}(90+A)+x \cos A \cot (90+A)=\sin (90+A) \text { then the value of } x \text { is }$
- A$\operatorname{cosec} A$
- B$\cot A$
- Correct answerC$\tan A$
- D$\sin A$
Solution
Using trigonometric identities for allied angles:
$\operatorname{cosec}(90^{\circ} + A) = \sec A$
$\cot(90^{\circ} + A) = -\tan A$
$\sin(90^{\circ} + A) = \cos A$
Substituting these into the given equation:
$\sec A + x \cos A (-\tan A) = \cos A$
$\dfrac{1}{\cos A} - x \cos A \left(\dfrac{\sin A}{\cos A}\right) = \cos A$
$\dfrac{1}{\cos A} - x \sin A = \cos A$
Rearranging the terms to solve for $x \sin A$:
$x \sin A = \dfrac{1}{\cos A} - \cos A$
$x \sin A = \dfrac{1 - \cos^2 A}{\cos A}$
$x \sin A = \dfrac{\sin^2 A}{\cos A}$
$x = \dfrac{\sin^2 A}{\cos A \sin A}$
$x = \dfrac{\sin A}{\cos A} = \tan A$
Answer: $\tan A$
$\operatorname{cosec}(90^{\circ} + A) = \sec A$
$\cot(90^{\circ} + A) = -\tan A$
$\sin(90^{\circ} + A) = \cos A$
Substituting these into the given equation:
$\sec A + x \cos A (-\tan A) = \cos A$
$\dfrac{1}{\cos A} - x \cos A \left(\dfrac{\sin A}{\cos A}\right) = \cos A$
$\dfrac{1}{\cos A} - x \sin A = \cos A$
Rearranging the terms to solve for $x \sin A$:
$x \sin A = \dfrac{1}{\cos A} - \cos A$
$x \sin A = \dfrac{1 - \cos^2 A}{\cos A}$
$x \sin A = \dfrac{\sin^2 A}{\cos A}$
$x = \dfrac{\sin^2 A}{\cos A \sin A}$
$x = \dfrac{\sin A}{\cos A} = \tan A$
Answer: $\tan A$
Hence, the correct answer is option: 'C'.