Question

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

• A. $2$
• B. ${\sin }^2\theta$
• C. ${\cos }^2\theta$
• D. $1$

Answer 1

The correct option is D $1$

$\csc \theta -\sin \theta =a^3$

$\Rightarrow \frac{1-{\sin }^2\theta }{\sin \theta }=a^3$

$\Rightarrow \frac{{\cos }^2\theta }{\sin \theta }=a^3-\left(1\right)$

$\sec \theta -\cos \theta =b^3$

$\Rightarrow \frac{1-{\cos }^2\theta }{\cos \theta }=b^3$

$\Rightarrow \frac{{\sin }^2\theta }{\cos \theta }=b^3-\left(2\right)$

$\left(1\right)÷\left(2\right)\Rightarrow \frac{{\cos }^3\theta }{{\sin }^3\theta }=\frac{a^3}{b^3}$

$\tan \theta =\frac{b}{a}-\left(3\right)$

$\left(1\right)\times \left(2\right)\Rightarrow \frac{{\cos }^2\theta }{\sin \theta }\times \frac{{\sin }^2\theta }{\cos \theta }=a^3{b}^3$

$\sin \theta .\cos \theta =a^3{b}^3-\left(4\right)$

$\left(3\right)\times \left(4\right)\Rightarrow {\sin }^2\theta =a^2{b}^4$

$\left(4\right)÷\left(3\right)\Rightarrow {\cos }^2\theta =a^4{b}^2$

$\left(3\right)\Rightarrow 1+\frac{b^2}{a^2}=1+{\tan }^2\theta ={\sec }^2\theta$

Given, $a^2{b}^2(a^2+b^2)=a^4{b}^2(1+\frac{b^2}{a^2})$

$={\cos }^2\theta \times {\sec }^2\theta =1$