Question

The value of $\tan hx$ if $5\sin hx-\cos hx=5$ is:

• A. $\frac{4}{5}$
• B. $-\frac{3}{5}$
• C. $\frac{5}{3}$
• D. $\frac{-5}{4}$

Answer 1

Answer: (A), (B)

The correct options are
A $\frac{4}{5}$
B $-\frac{3}{5}$

We have $5\sin hx-\cos hx=5$
$\Rightarrow 5\sin hx-5=\cos hx$
$\Rightarrow 5(\sin hx-1)=\cos hx$
Squaring both sides, we get
$\Rightarrow 25{(\sin hx-1)}^2={\left(\cos hx\right)}^2$
$\Rightarrow 25[\sin h^{2}x+1-2\sin hx]=\cos h^{2}x$
We know that $\cos h^{2}x-\sin h^{2}x=1$
$\Rightarrow 25\sin h^{2}x+25-50\sin hx=1+\sin h^{2}x$
$\Rightarrow 24\sin h^{2}x-50\sin hx+24=0$
On factorising, we get
$(3\sin hx-4)(4\sin hx-3)=0$
whence $\sin hx=\frac{4}{3}$ or $\frac{3}{4}$
$\therefore \cos hx=\sqrt{1+\sin h^{2}x}=\sqrt{1+{\left(\frac{4}{3}\right)}^2}=\frac{5}{3}$
and $\cos hx=\sqrt{1+\sin h^{2}x}=\sqrt{1+{\left(\frac{3}{4}\right)}^2}=\pm \frac{5}{4}=-\frac{5}{4}$ $(\because \cos hx=\frac{5}{4})$ does not satisfy the above equation.
Hence $\tan hx=\frac{4}{5}$ or $\frac{-3}{5}$