Plz question 27 fast - Maths - Trigonometric Ratios

Question

Plz question 27 fast

Ankita Agarwal · Accepted Answer

Dear Student,

solving this question by Taking m as p
So ,

given: $\cos e c A + \cot A = p$ ..............(1)

$(\cos e c A + \cot A)^{2} = p^{2} \frac{(\cos e c A + \cot A)^{2}}{1} = \frac{p^{2}}{1}$

using componendo and dividendo:

$\frac{(\cos e c A + \cot A)^{2} - 1}{(\cos e c A + \cot A)^{2} + 1} = \frac{p^{2} - 1}{p^{2} + 1} \frac{\cos e c^{2} A + \cot^{2} A + 2 \cos e c A \cot A - 1}{\cos e c^{2} A + \cot^{2} A + 2 \cos e c A \cot A + 1} = \frac{p^{2} - 1}{p^{2} + 1} \frac{p^{2} - 1}{p^{2} + 1} = \frac{(\cos e c^{2} A - 1) + \cot^{2} A + 2 \cos e c A \cot A}{\cos e c^{2} A + (1 + \cot^{2} A) + 2 \cos e c A \cot A}$

$= \frac{\cot^{2} A + \cot^{2} A + 2 \cos e c A \cot A}{\cos e c^{2} A + \cos e c^{2} A + 2 \cos e c A \cot A} = \frac{2 \cot^{2} A + 2 \cos e c A \cot A}{2 \cos e c^{2} A + 2 \cos e c A \cot A} = \frac{2 \cot A (\cot A + \cos e c A)}{2 \cos e c A (\cos e c A + \cot A)} = \frac{\cot A}{\cos e c A} = \frac{\cos A}{\sin A * 1 / \sin A} = \cos A$

thus

$\cos A = \frac{p^{2} - 1}{p^{2} + 1}$

Regards