If $\tan x=\sqrt{\frac{a}{b}}$ where a,b are positive real numbers and x is in 1st quadrant then find the value of $\sin x\sec^7x+\cos x\csc^7x$ The answer is $\frac{(a+b)^3(a^4+b^4)}{(ab)^{\frac{7}{2}}}$. I just want to now how to do it. 
 A: you say $$\tan t = \sqrt{a/b}, 0 < t < \pi/2 \to y = \sin t = \sqrt{\frac a{a+b}}, x = \cos t = \sqrt{\frac b{a+b}}$$
then $$\begin{align}\sin t \sec^7 t + \cos t \csc ^7 t &= \frac y{x^7} + \frac x{y^7} \\
&= \frac{x^8+y^8}{(xy)^7} \\
&= \frac{(a+b)^3(a^4+b^4)}{(ab)^{7/2}}\end{align}$$
A: We very well know that $$\color {blue}{\sec^2 x-\tan^2 x=1}$$ $$\implies  \sec^2 x=1+\tan^2 x=1+\left(\sqrt{\frac{a}{b}}\right)^2=1+\frac{a}{b}=\frac{a+b}{b}$$ $$\implies \sec x=\pm \sqrt{\frac{a+b}{b}}$$ But x lies in the first quadrant i.e. $\color{blue}{0<x<\frac{\pi}{2}}$ hence $\sin x$, $\cos x$, $\csc x$ & $\sec x$ all have positive values in the I-quadrant, thus we have $$\color {blue}{ \sec x= \sqrt{\frac{a+b}{b}}}$$ $$\implies \cos x=\frac{1}{\sec x}=\frac{1}{\sqrt{\frac{a+b}{b}}}=\sqrt{\frac{b}{a+b}}$$  $$\implies \sin x=\frac{\tan x}{\sec x}=\frac{\sqrt{\frac{a}{b}}}{\sqrt{\frac{a+b}{b}}}=\sqrt{\frac{a}{a+b}}$$ $$\implies \csc x=\frac{1}{\sin x}=\sqrt{\frac{a+b}{a}}$$ 
 Now substituting the above values, we have $$\sin x\sec^7 x+\cos x\csc^7 x=\sqrt{\frac{a}{a+b}}\left(\sqrt{\frac{a+b}{b}}\right)^7+\sqrt{\frac{b}{a+b}}\left(\sqrt{\frac{a+b}{a}}\right)^7 $$ $$=\frac{(a+b)^3}{b^3} \sqrt{\frac{a}{a+b}}\sqrt{\frac{a+b}{b}}+\frac{(a+b)^3}{a^3}\sqrt{\frac{b}{a+b}}\sqrt{\frac{a+b}{a}}$$ $$=\frac{(a+b)^3}{b^3}\sqrt{\frac{a}{b}}+\frac{(a+b)^3}{a^3}\sqrt{\frac{b}{a}} $$ $$=(a+b)^3\left[\frac{a^{\frac{1}{2}}}{b^{\frac{7}{2}}}+\frac{b^{\frac{1}{2}}}{a^{\frac{7}{2}}}\right]$$ $$=(a+b)^3\left[\frac{a^4+b^4}{a^{\frac{7}{2}}b^{\frac{7}{2}}}\right]$$$$=\color{purple}{\frac{(a+b)^3(a^4+b^4)}{(ab)^{\frac{7}{2}}}}$$ 
