Question

Can you solve question number three??

Answer 1

Here I'm using the following:

$1.\ \ \csc(90\textdegree-x)=\sec x\quad\quad\text{OR}\quad\quad\sec(90\textdegree-x)=\csc x\\\\2.\ \ \cos(90\textdegree-x)=\sin x\\\\3.\ \ \sin^2x+\cos^2x=1\\\\4.\ \ \sec^2x-\tan^2x=1\\\\5.\ \ \sin x\csc x=1\\\\\\6.\ \ \tan30\textdegree=\dfrac{1}{\sqrt3}$

So,

$\begin{aligned}&\dfrac{\csc^2(90\textdegree-\theta)-\tan^2\theta}{4(\cos^240\textdegree+\cos^250\textdegree)}-\dfrac{2\tan^230\textdegree\sec^252\textdegree\sin^238\textdegree}{3(\csc^270\textdegree-\tan^220\textdegree)}\end{aligned}$

$\Longrightarrow\ \ &\dfrac{\sec^2\theta-\tan^2\theta}{4(\cos^2(90\textdegree-50\textdegree)+\cos^250\textdegree)}-\dfrac{2\left(\dfrac{1}{\sqrt3}\right)^2\sec^2(90\textdegree-38\textdegree)\sin^238\textdegree}{3(\csc^2(90\textdegree-20\textdegree)-\tan^220\textdegree)}$

$\Longrightarrow\ \ \dfrac{1}{4(\sin^250\textdegree+\cos^250\textdegree)}-\dfrac{\dfrac{2}{3}\cdot\csc^238\textdegree\sin^238\textdegree}{3(\sec^220\textdegree-\tan^220\textdegree)}\\\\\\\implies\ \ \dfrac{1}{4}-\dfrac{2}{9}\ \implies\ \boxed{\dfrac{1}{36}}$