Question

Evaluate :

$si{n}^2{34}^{\circ }+si{n}^2{56}^{\circ }+2tan{18}^{\circ }tan{72}^{\circ }-co{t}^{2}30\circ$

Answer 1

sin (90-x) = cos x and tan (90 -x) = tan x.
Therefore, sin 56 = sin (90 - 34) = cos 34. Tan 72 = cot 18,
The value of $cot30=\frac{31}{2}$

With the above substitutions, the given expression tuns to:
$(si{n}^2{18}^0+co{s}^2{18}^0)+2ta{n}^2{18}^0\ast cot18-3$
= 1 + 2 tan 18 -3
= 2 (tan 18 - 1) = 2( 0.3249 - 1) = -1.35 nearly.

Note: If the 3rd term in the given expression is $2ta{n}^2{18}^{0}ta{n}^2{72}^0$ instead of, $2ta{n}^2{18}^{0}tan{72}^0$ then the answer would be zero.