CUET Preparation Today
What is the value of $\frac{3cos62^o}{sin28^o}-\frac{2tan34^o}{cot56^o}$? |
3 1 5 4 |
1 |
The concept we use here is :- sinθ = cos ( 90º - θ ) & tanθ = cot ( 90º - θ ) Now, \(\frac{3 cos62º}{sin28º}\) - \(\frac{2 tan34º}{cot56º}\) = \(\frac{3 sin28º}{sin28º}\) - \(\frac{2 tan34º}{tan34º}\) = 3 -2 = 1 |
