RD Chapter 9 Trigonometric Ratios of Multiple and Submultiple Angles Ex 9.1 Solutions
Question - 21 : - cos6 x– sin6 x = cos 2x (1 – 1/4 sin2 2x)
Answer - 21 : -
Let us consider LHS:
cos6 x – sin6 x
We know, (a + b) 2 = a2 +b2 + 2ab
a3 – b3 = (a – b) (a2 +b2 + ab)
So,
cos6 x – sin6 x = (cos2 x)3 –(sin2 x)3
= (cos2 x – sin2 x)(cos4 x + sin4 x + cos2 x sin2 x)
We know, cos 2x = cos2 x – sin2 x
So,
= cos 2x [(cos2 x) 2 +(sin2 x) 2 + 2 cos2 x sin2 x– cos2 x sin2 x]
= cos 2x [(cos2 x) 2 +(sin2 x) 2 – 1/4 × 4 cos2 xsin2 x]
We know, sin2 x + cos2 x= 1
So,
= cos 2x [(1)2 – 1/4 × (2 cos x sin x) 2]
We know, sin 2x = 2 sin x cos x
So,
= cos 2x [1 – 1/4 × (sin 2x) 2]
= cos 2x [1 – 1/4 × sin2 2x]
= RHS
Hence proved.
Question - 22 : - tan (π/4 + x) + tan (π/4 – x) = 2sec 2x
Answer - 22 : -
Let us consider LHS:
tan (π/4 + x) + tan (π/4 – x)
We know,
tan (A+B) = (tan A + tan B)/(1- tan A tan B)
tan (A-B) = (tan A – tan B)/(1+ tan A tan B)
So,
