Question

If 4 tan $\theta =3$, evaluate $\left(\frac{4sin\theta -cos\theta +1}{4sin\theta +cos\theta -1}\right)$
OR
If tan 2A = cot $(A-{18}^o)$, where 2A is an acute angle, find the value of A.

Answer 1

Given, 4 tan $\theta =3,$
$\Rightarrow tan\theta =\frac{3}{4}(=\frac{P}{B})$

P = 3K, B = 4K,
Now, H = $\sqrt{P^2+B^2}$
$=\sqrt{(3K{)}^2+(4K{)}^2}$
$=\sqrt{9K^2+16K^2}$
$=\sqrt{25K^2}$
$\Rightarrow H=5K$
$\therefore sin\theta =\frac{P}{H}=\frac{3K}{5K}=\frac{3}{5}$
and $cos\theta =\frac{B}{H}=\frac{4K}{5K}=\frac{4}{5}$
Now, $\frac{4sin\theta -cos\theta +1}{4sin\theta +cos\theta -1}=\frac{4\times \frac{3}{5}-\frac{4}{5}+1}{4\times \frac{3}{5}+\frac{4}{5}-1}$

$=\frac{(\frac{12}{5}-\frac{4}{5}+1)}{(\frac{12}{5}+\frac{4}{5}-1)}$

$=\frac{\left(\frac{12-4+5}{5}\right)}{\left(\frac{12+4-5}{5}\right)}$

$=\frac{13/5}{11/5}$
$=\frac{13}{11}$
OR
Given, tan 2A = cot $(A-{18}^o)$
$\Rightarrow cot({90}^o-2A)=cot(A-{18}^o)[\because tan\theta =cot({90}^o-\theta \left)\right]$
$\Rightarrow {90}^o-2A=A-{18}^o$
$\Rightarrow {90}^o+{18}^o=A+2A$
$\Rightarrow {180}^o=3A$
$\Rightarrow A=\frac{{108}^o}{3}$
$\Rightarrow A={36}^o$