Find the value of $$\cot 19°$$​  $$\bigg(cot71° cos²21°+\frac{1}{tan71° sec²69°}\bigg)$$​​

Correct option is D

Given:

​$$\cot19^o \bigg(cot71° cos²21°+\frac{1}{tan71° sec²69°}\bigg)$$​​

Formula Used:

​$$\tan \theta = \cot(90^o - \theta)$$​​

​$$\cot \theta = \tan(90^o - \theta)$$​​

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

​$$\cos \theta = \frac{1}{\sec \theta}$$​​

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

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

Solution:

​$$\cot19^o \bigg(cot71° cos²21°+\frac{1}{tan71° sec²69°}\bigg)$$​​

= $$\bigg((\cot19^o)cot71° cos²21°+\frac{(\cot19^o)}{tan71° sec²69°}\bigg)$$​​

=$$\bigg((\tan(90^o - 19^o))cot71° cos²21°+\frac{(\tan(90^o - 19^o))}{tan71° sec²69°}\bigg)$$​​

=$$\bigg((\tan71^ocot71° cos²21°+\frac{\tan71^o}{tan71° sec²69°}\bigg)$$​​

=$$\bigg((\frac{1}{\ cot71^o})cot71° cos²21°+\frac{1}{ sec²69°}\bigg)$$​​

= $$\bigg( cos²21°+\frac{1}{ sec²69°}\bigg)$$​​

= $$( cos²21°+cos²69°)$$​​

= $$( cos²21°+\sin^2(90-69)°)$$​​

= $$( cos²21°+\sin^21°)$$​​

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

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