Question

If $cosec\theta -sin\theta =a^3$ and $sec\theta -cos\theta =b^3$, then $a^2{b}^2(a^2+b^2)=$

• A. 1
• B. -1
• C. 2
• D. None of these

Answer 1

The correct option is A

1

$\Rightarrow cosec\theta -sin\theta =a^3$
$a^3=(\frac{1}{sin\theta }-sin\theta )$
$a^3=\left(\frac{1-si{n}^2\theta }{sin\theta }\right)$
$a^3=\frac{co{s}^2\theta }{sin\theta }$
$(\because si{n}^2\theta +co{s}^2\theta =1)$

$\Rightarrow sec\theta -cos\theta =b^3$
$b^3=sec\theta -cos\theta$
$b^3=\frac{1}{cos\theta }-cos\theta$
$b^3=\frac{1-co{s}^2\theta }{cos\theta }$
$b^3=\frac{si{n}^2\theta }{cos\theta }$
$(\because si{n}^2\theta +co{s}^2\theta =1)$

$⟹a^3\times b^3=\frac{co{s}^2\theta }{sin\theta }\times \frac{si{n}^2\theta }{cos\theta }$

$\Rightarrow a^3{b}^3=sin\theta .cos\theta$ and $\frac{a^3}{b^3}=\frac{co{s}^3\theta }{si{n}^3\theta }$

Hence, $\frac{a}{b}=\frac{cos\theta }{sin\theta }$

Now, multiply $\frac{ab}{ab}$ to $a^2{b}^2(a^2+b^2)$

$\Rightarrow a^2{b}^2(a^2+b^2)=a^3{b}^3(\frac{a}{b}+\frac{b}{a})$

Now substituting the values, we get

$=sin\theta .cos\theta [\frac{cos\theta }{\sin \theta }+\frac{sin\theta }{\cos \theta }]$

$=(sin\theta .cos\theta \times \frac{cos\theta }{\sin \theta })+(\frac{sin\theta }{\cos \theta }\times sin\theta .cos\theta )$

$=co{s}^2\theta +si{n}^2\theta$

$=1(\because si{n}^2\theta +co{s}^2\theta =1$)