Question

(i) If $\cos 3A=\sin (A-{34}^0)$, where A is an acute angle, find the value of A.
(ii) Prove the following identity, where the angles involved are acute angles for which the expression is define.
$\frac{1+{\cot }^{2}A}{1+{\tan }^{2}A}={\left(\frac{1-\cot A}{1-\tan A}\right)}^2$

Answer 1

(i) $\cos 3A=\sin (A-34)$

$=\cos (90-A+34)$

$=\cos (124-A)$

So, $124-A=2n\pi \pm 3A$

For $+ve$

$124-A=2n\pi +3A$

$4A=124-2n\pi$

as $A$ is acute $n=0$

$A=\frac{124}{4}=31°$

For $-ve$

$124-A=2n\pi -3A$

$124+2A=2n\pi$

$2A=2n\pi -124$

as $A$ is acute, no $n$ exists

$A=31°$

(ii)$\frac{1+{\cot }^{2}A}{1+{\tan }^{2}A}=\frac{{\csc }^{2}A}{{\sec }^{2}A}={\cot }^{2}A$

${\left(\frac{1-\cot A}{1-\tan A}\right)}^2={\left(\frac{1-\frac{1}{\tan A}}{1-\tan A}\right)}^2={\left(\frac{\tan A-1}{1-\tan A}\right)}^2\times \frac{1}{{\tan }^{2}A}={\cot }^{2}A$

So, $\frac{1+{\cot }^{2}A}{1+{\tan }^{2}A}={\left(\frac{1-\cot A}{1-\tan A}\right)}^2$