Question

If $\tan \theta =\sqrt{\frac{a}{b}}$ where $a,b$ are positive real and $\theta \in 1^{st}$ quadrant then the value of $\sin \theta {\sec }^7\theta +\cos \theta {\csc }^7\theta$ is

• A. $\frac{{(a+b)}^3(a^4+b^4)}{{\left(ab\right)}^{7/2}}$
• B. $\frac{{(a+b)}^3(a^4-b^4)}{{\left(ab\right)}^{7/2}}$
• C. $\frac{{(a+b)}^3(b^4-a^4)}{{\left(ab\right)}^{7/2}}$
• D. $-\frac{{(a+b)}^3(a^4+b^4)}{{\left(ab\right)}^{7/2}}$

Answer 1

The correct option is A $\frac{{(a+b)}^3(a^4+b^4)}{{\left(ab\right)}^{7/2}}$

$tan\theta =\sqrt{\frac{a}{b}}$

$se{c}^2\theta =1+ta{n}^2\theta$

$se{c}^2\theta =\frac{a+b}{b}$

$cot\theta =\sqrt{\frac{b}{a}}$

$cose{c}^2\theta =1+co{t}^2\theta =\frac{a+b}{a}$

$sin\theta se{c}^7\theta +cos\theta cs{c}^7\theta =tan\theta (se{c}^2\theta {)}^3+cot\theta (cose{c}^2\theta {)}^3$

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

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

$=\frac{(a+b{)}^3(a^4+b^4)}{(ab{)}^{\frac{7}{2}}}$