The value of $$\lparen\frac{\text{sin}\space29°}{\text{cos}\space61°}-4(\text{cos}^231°+\text{cos}^259°)+\frac{1}{\sqrt3}\space\text{tan}\space27°\space\text{sin}\space60°\space\text{tan}\space63°\rparen$$​ is:

Correct option is A

Given:

​$$\left( \frac{\sin 29^\circ}{\cos 61^\circ} - 4(\cos^2 31^\circ + \cos^2 59^\circ) + \frac{1}{\sqrt{3}} \tan 27^\circ \sin 60^\circ \tan 63^\circ \right)$$​​

Formula Used:

​$$\sin \theta = \cos(90^\circ - \theta)$$​​

​$$\cos^2 \theta + \sin^2 \theta = 1$$ 

tan $$\theta = \frac 1{\cot \theta}$$ ​

Solution:

​$$\left( \frac{\sin 29^\circ}{\cos 61^\circ} - 4(\cos^2 31^\circ + \cos^2 59^\circ) + \frac{1}{\sqrt{3}} \tan 27^\circ \sin 60^\circ \tan 63^\circ \right)$$​​

= $$\frac{\sin 29^\circ}{\sin 29^\circ}$$$$- 4(\cos^2 31^\circ + \sin^2 31^\circ) +$$​ $$\frac{1}{\sqrt{3}} \cdot \tan 27^\circ \cdot \sin 60^\circ \cdot \cot 27^\circ$$

= 1- 4( 1) + $$\frac{1}{\sqrt{3}} \times \frac{\sqrt{3}}{2} \times 1$$​​​

= 1 - 4 $$+ \frac{1}{2} = -3 + \frac{1}{2} = -\frac{5}{2}$$