Question

A right $\Delta ABC$ right angled at B and $AC=\sqrt{10}$ and AB = 1 and $\underset{–}{ACB}=\sec \theta {\cot }^2\theta$ is

• A. 1
• B. 19
• C. $\frac{3}{10}$
• D. $\frac{\sqrt{19}}{10}$

Answer 1

The correct option is B 19

Here, AB is perpendicular(P), BC is base(B) and AC is hypotenuse(H).
$BC=\sqrt{A{C}^2-A{B}^2}=\sqrt{10-1}=3$
$\therefore sin\theta =\frac{P}{H}=\frac{1}{\sqrt{10}}$
$\Rightarrow cosec\theta =\sqrt{10}$ ...(i)
Also, $tan\theta =\frac{P}{B}=\frac{1}{3}$
$\Rightarrow cot\theta =3$ ...(ii)
On squaring and adding (i) and (ii), we get
$cose{c}^2\theta +co{t}^2\theta =10+9=19$
Option B is correct.