RD Chapter 5 Trigonometric Functions Ex 5.1 Solutions
Question - 1 : - Prove the following identities:sec4 x – sec2 x= tan4 x + tan2 x
Answer - 1 : -
Let us consider LHS: sec4 x– sec2 x
(sec2 x)2 –sec2 x
By using the formula, sec2 θ = 1 + tan2 θ.
(1 + tan2 x) 2 –(1 + tan2 x)
1 + 2tan2 x + tan4 x– 1 – tan2 x
tan4 x + tan2 x
= RHS
∴LHS = RHS
Hence proved.
Question - 2 : - sin6 x + cos6 x= 1 – 3 sin2 x cos2 x
Answer - 2 : -
Let us consider LHS: sin6 x+ cos6 x
(sin2 x) 3 +(cos2 x) 3
By using the formula, a3 + b3 =(a + b) (a2 + b2 – ab)
(sin2 x + cos2 x)[(sin2 x) 2 +(cos2 x) 2 –sin2 x cos2 x]
By using the formula, sin2 x + cos2 x= 1 and a2 + b2 = (a + b) 2 –2ab
1 × [(sin2 x + cos2 x) 2 – 2sin2 xcos2 x – sin2 x cos2 x
12 –3sin2 x cos2 x
1 – 3sin2 x cos2 x
= RHS
∴LHS = RHS
Hence proved.
Question - 3 : - (cosec x – sin x) (sec x – cos x) (tan x + cot x) = 1
Answer - 3 : -
Let us consider LHS: (cosec x – sin x) (sec x – cos x)(tan x + cot x)
By using the formulas
cosec θ = 1/sin θ;
sec θ = 1/cos θ;
tan θ = sin θ / cos θ;
cot θ = cos θ / sin θ
Now,
1 = RHS
∴ LHS = RHS
Hence proved.
Question - 4 : - cosec x (sec x – 1) – cot x (1 – cos x) = tan x – sin x
Answer - 4 : -
Let us consider LHS: cosec x (sec x – 1) – cot x (1 – cos x)
By using the formulas
cosec θ = 1/sin θ;
sec θ = 1/cos θ;
tan θ = sin θ / cos θ;
cot θ = cos θ / sin θ
Now,
By using the formula, 1 – cos2x = sin2x;
= RHS
∴ LHS = RHS
Hence Proved.
Question - 5 : - 
Answer - 5 : -
Let us consider the LHS:
By using the formula,
cosec θ = 1/sin θ;
sec θ = 1/cos θ;
Now,
sin x
= RHS
∴ LHS = RHS
Hence Proved.
Question - 6 : - 
Answer - 6 : -
Let us consider the LHS:
By using the formula,
tan θ = sin θ / cos θ;
cot θ = cos θ / sin θ
Now,
By using the formula, a3 – b3 =(a – b) (a2 + b2 + ab)
By using the formula,
cosec θ = 1/sin θ,
sec θ = 1/cos θ;
cosec x × sec x + 1
sec x cosec x + 1
=RHS
∴ LHS = RHS
Hence Proved.
Answer - 7 : - Let us consider LHS:
By using the formula a3 ± b3 =(a ± b) (a2 + b2∓ ab)
We know, sin2x + cos2x = 1.
1 – sinx cosx + 1 + sinx cosx
2
= RHS
∴ LHS = RHS
Hence Proved.
Answer - 8 : -
Let us consider LHS:
(sec x sec y + tan x tan y)2 – (sec xtan y + tan x sec y)2
Expanding the above equation we get,
[(sec x secy)2 + (tan x tan y)2 + 2 (sec x sec y) (tan xtan y)] – [(sec x tan y)2 + (tan x sec y)2 + 2(sec x tan y) (tan x sec y)] [sec2 x sec2 y +tan2 x tan2 y + 2 (sec x sec y) (tan x tan y)]– [sec2 x tan2 y + tan2 x sec2 y+ 2 (sec2 x tan2 y) (tan x sec y)]
sec2 x sec2 y – sec2 xtan2 y + tan2 x tan2 y – tan2 xsec2 y
sec2 x (sec2 y – tan2 y)+ tan2 x (tan2 y – sec2 y)
sec2 x (sec2 y – tan2 y)– tan2 x (sec2 y – tan2 y)
We know, sec2 x – tan2 x= 1.
sec2 x × 1 – tan2 x ×1
sec2 x – tan2 x
1 = RHS
∴ LHS = RHS
Hence proved.
Question - 9 : - 
Answer - 9 : -
Let us Consider RHS:
= LHS
∴ LHS = RHS
Hence Proved.
Question - 10 : - 
Answer - 10 : -
Let us consider LHS:
By using the formulas,
1 + tan2x = sec2x and 1 + cot2x= cosec2x
= RHS
∴ LHS = RHS
Hence Proved.