Question

sin square 34 degree + sin square 56 degree + 2 tan 18 degrees tan72 degree minus cos square 30 degree

Answer 1

hey dear!!
here is your answer.
hope it helps u✌

Answer 2

The value of $\sin^2 34+\sin^2 56+2\tan 18\tan 72-\cos^2 30=\dfrac{9}{4}$.

Step-by-step explanation:

We have,

$\sin^2 34+\sin^2 56+2\tan 18\tan 72-\cos^2 30$

To find, the value of $\sin^2 34+\sin^2 56+2\tan 18\tan 72-\cos^2 30=?$

$=\sin^2 34+\sin^2 (90-34)+2\tan 18\tan (90-18)-\cos^2 30$

Using trigonometric identity,

$\sin(90-A)=\cos A$ and $\tan(90-A)=\cot A$

$=\sin^2 34+\cos^2 34+2\tan 18\cot 18-\cos^2 30$

Using trigonometric identity,

$\sin^2A+\cos^2 A=1$

$=1+2\tan 18\cot 18-\cos^2 30$

Using trigonometric identity,

$\tan A.\cot A=1$

$=1+2(1)-(\dfrac{\sqrt{3}}{2})^2$

$=3-\dfrac{3}{4}=\dfrac{12-3}{4}$

$=\dfrac{9}{4}$

Hence, the value of $\sin^2 34+\sin^2 56+2\tan 18\tan 72-\cos^2 30=\dfrac{9}{4}$.