Show that:
(i) tan 48° tan 23° tan 42° tan 67° = 1
(ii) cos 38° cos 52° - sin 38° sin 52° = 0

Solution:

We will be using the trigonometric ratios of complementary angles to solve the given question.

sin (90° - θ) = cos θ

tan (90° - θ) = cot θ

(i) Taking L.H.S, tan 48° tan 23° tan 42° tan 67°

Since tan (90° - θ) = cot θ

tan 48° tan 23° tan 42° tan 67° = tan (90° - 42°) tan(90° - 67°) tan 42° tan 67°

= cot 42° cot 67° tan 42° tan 67°

= (cot 42° tan 42°)(cot 67° tan 67°)

= (1/tan 42° × tan 42°)(1/tan 67° × tan 67°)

= 1 × 1

= 1

= R.H.S

Hence, tan 48°tan 23°tan 42° tan 67° = 1

(ii) Taking L.H.S, cos 38° cos 52° - sin 38° sin 52°

sin(90° - θ) = cos θ

cos 38° cos 52° - sin 38° sin 52° = cos 38° cos 52° - sin (90° - 52°) sin (90° - 38°)

= cos 38° cos 52° - cos 52° cos 38°

= 0

= R.H.S

Hence, cos 38° cos 52° - sin 38° sin 52° = 0

☛ Check: NCERT Solutions for Class 10 Maths Chapter 8

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Video Solution:

Show that: (i) tan 48° tan 23° tan 42° tan 67° = 1 (ii) cos 38° cos 52° - sin 38° sin 52° = 0

Maths NCERT Solutions Class 10 Chapter 8 Exercise 8.3 Question 2

Summary:

It is proved that, (i) tan 48° tan 23° tan 42° tan 67° = 1, (ii) cos 38° cos 52° − sin 38° sin 52° = 0

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