cos (arctan (frac{pi }{2}))-sin (arc cot (frac{pi }{4}))cos(arctan(π2))−sin(arccot(π4))
We know,
cos (arctan (frac{pi }{2}))cos(arctan(π2))
using the following identity,
cos (arctan (x))=frac{1}{sqrt{1+x^2}}cos(arctan(x))=1√1+x2
=frac{1}{sqrt{1+(frac{pi }{2})^2}}=1√1+(π2)2
Refining it,
=\frac{2}{\sqrt{4+\pi ^2}}=2√4+π2
Also,
sin (arc cot (frac{pi }{4}))sin(arccot(π4))
using the following identity,
sin (arc cot (x))=frac{1}{sqrt{1+x^2}}sin(arccot(x))=1√1+x2
sin (arc cot (x))=frac{1}{sqrt{1+x^2}}sin(arccot(x))=1√1+x2
Refining it,
=frac{4}{sqrt{16+pi ^2}}=4√16+π2
Finally,
=frac{2}{sqrt{pi ^2+4}}-frac{4}{sqrt{pi ^2+16}}=2√π2+4−4√π2+16
