Practicing Success

Target Exam

CUET

Subject

General Test

Chapter

Quantitative Reasoning

Topic

Trigonometry

Question:

The value of $\frac{\tan \left(45^{\circ}-\alpha\right)}{\cot \left(45^{\circ}+\alpha\right)}-\frac{\left(\cos 19^{\circ}+\sin 71^{\circ}\right)\left(\sec 19^{\circ}+{cosec} 71^{\circ}\right)}{\tan 12^{\circ} \tan 24^{\circ} \tan 66^{\circ} \tan 78^{\circ}}$ is:

Options:

-3

0

-2

2

Correct Answer:

-3

Explanation:

\(\frac{tan (45º-α) }{cot (45º+α) }\) - \(\frac{(cos19º +sin71º).(sec19º+cosec71º) }{tan12º.tan24º.tan66º.tan78º }\)

Using ,

tan (90º - A ) = cotA

sin(90º-A) = cosA

cosec(90º-A)= secA

cot (90º-A) = tanA

= \(\frac{tan (90º - 45º+α) }{cot (45º+α) }\) - \(\frac{(sin71º +sin71º).(cosec71º+cosec71º) }{cot78º.cot66º.tan66º.tan78º }\)

= \(\frac{cot( 45º+α) }{cot (45º+α) }\) - \(\frac{( 2 sin71º ).( 2cosec71º) }{1 }\)

= 1 - 4

= -3