Question

If $cosec\theta -sin\theta =a^3$ and $sec\theta -cos\theta =b^3$ then, the value of $a^2{b}^2(a^2+b^2)$ is __________ .

• A. 1
• B. -1
• C. 2
• D. -2

Answer 1

The correct option is A 1

Given: $cosec\theta -sin\theta =a^3$ and $sec\theta -cos\theta =b^3$

$\Rightarrow cosec\theta -sin\theta =a^3$
We know that, $cosec\theta =\frac{1}{sin\theta }.$
$a^3=(\frac{1}{sin\theta }-sin\theta )$
$a^3=\left(\frac{1-si{n}^2\theta }{sin\theta }\right)$

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

Given, $b^3=sec\theta -cos\theta$
We know that, $sec\theta =\frac{1}{cos\theta }.$
$b^3=\frac{1}{cos\theta }-cos\theta$
$b^3=\frac{1-co{s}^2\theta }{cos\theta }$

$b^3=\frac{si{n}^2\theta }{cos\theta }....\left(ii\right)(\because 1-co{s}^2\theta =si{n}^2\theta )$

From (i) and (ii),
$a^3\times b^3=\frac{co{s}^2\theta }{sin\theta }\times \frac{si{n}^2\theta }{cos\theta }$
By cancelling out the common terms, we get,
$a^3\times b^3=sin\theta \times cos\theta$ 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 $a^2{b}^2(a^2+b^2)$ we get,

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

Now substituting the values, we get

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

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

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