Question

Prove the following trigonometric equation if $a\cos ecA = p$ and $b\cot A = q$
$\dfrac{{{p^2}}}{{{a^2}}} - \dfrac{{{q^2}}}{{{b^2}}} = 1$.

Answer 1

Hint: In order to solve this problem use the formulas $\cos ecx = \dfrac{1}{{\sin x}}{\text{ and }}\cot x = \dfrac{{\cos x}}{{\sin x}}$. Using these and solving will provide you the right answer.

Complete step-by-step answer:
The given equation is $\dfrac{{{p^2}}}{{{a^2}}} - \dfrac{{{q^2}}}{{{b^2}}} = 1$………..(1)
It is also given that $a\cos ecA = p$ and $b\cot A = q$.
On putting the value of p and q in (1) we get,
The equation as:
$\dfrac{{{a^2}\cos e{c^2}A}}{{{a^2}}} - \dfrac{{{b^2}{{\cot }^2}A}}{{{b^2}}} = 1 \\\ \cos e{c^2}A - {\cot ^2}A = 1 \\\ \dfrac{1}{{{{\sin }^2}A}} - \dfrac{{{{\cos }^2}A}}{{{{\sin }^2}A}} = \dfrac{{1 - {{\cos }^2}A}}{{{{\sin }^2}A}} = \dfrac{{{{\sin }^2}A}}{{{{\sin }^2}A}} = 1\,\,\,\,\,\,\,\,\,\,\,\,({\text{Since 1 - }}{\cos ^2}x = {\sin ^2}x) \\\$
Hence, it is proved that LHS and RHS both are equal.

Note: In this problem you need to solve the equation by putting the values given in the question and using the formulas $\cos ec x = \dfrac{1}{{\sin x}}{\text{ and }}\cot x = \dfrac{{\cos x}}{{\sin x}}$ to get the answer to this problem. But we can substitute the value of a as well and we can directly prove using the identity $\cos e{c^2}A - {\cot ^2}A = 1$.