Question

sin 35°×cos 55°+cos 35°×sin 55°/cosec square 10°-- tan square 80°

Answer 1

thanks.......................

Answer 2

$\dfrac{\sin 35\cos 55 +\cos 35\sin 55}{\csc^2 10-\tan^2 80}=1$

Step-by-step explanation:

We have,

$\dfrac{\sin 35\cos 55 +\cos 35\sin 55}{\csc^2 10-\tan^2 80}$

To find, $\dfrac{\sin 35\cos 55 +\cos 35\sin 55}{\csc^2 10-\tan^2 80}=?$

∴ $\dfrac{\sin 35\cos 55 +\cos 35\sin 55}{\csc^2 10-\tan^2 80}$

$=\dfrac{\sin (35+55)}{\csc^2 10-\tan^2 (90-10)}$

[ ∵ $\sin (A+B)=\sin A\cos B +\cos A\sin B$]

$=\dfrac{\sin 90}{\csc^2 10-\cot^2 10}$

$=\dfrac{1}{1}$

[ ∵ $\sin 90=1$ and $\csc^2 A-\cot^2 A=1$]

= 1

Hence, $\dfrac{\sin 35\cos 55 +\cos 35\sin 55}{\csc^2 10-\tan^2 80}=1$