Show that [cos² (45° + θ) + cos² (45° - θ)]/ [tan (60° + θ) tan (30° - θ)] = 1

Solution:

Consider

LHS = [cos² (45° + θ) + cos² (45° - θ)]/ [tan (60° + θ) tan (30° - θ)]

We know that

cos A = sin (90° - A)

tan A = cot (90° - A)

tan A cot A = 1

By substituting it we get

= [cos² (45° + θ) + sin² (90°

- 45° + θ)]/ [tan (60° + θ) cot (90° - 30° + θ)]

So we get

= [cos² (45° + θ) + sin² (45° + θ)]/ [tan (60° + θ) cot (60° + θ)]

Here cos²  A + sin²  A = 1 and tan A cot A = 1

By substituting it

= 1/1

= 1

= RHS

Therefore, it is proved.

✦ Try This: Prove that sec (70° - θ) = cosec (20° + θ)

☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 8

NCERT Exemplar Class 10 Maths Exercise 8.3 Problem 13

Show that [cos² (45° + θ) + cos² (45° - θ)]/ [tan (60° + θ) tan (30° - θ)] = 1

Summary:

It is shown that [cos² (45° + θ) + cos² (45° - θ)]/ [tan (60° + θ) tan (30° - θ)] = 1

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