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RD Chapter 5- Trigonometric Functions Ex-5.1 Interview Questions Answers

Question 1 : Prove the following identities:

secx – secx= tanx + tanx

Answer 1 :

Let us consider LHS: secx– secx

(secx)2 –secx

By using the formula, sec2 θ = 1 + tan2 θ.

(1 + tanx) 2 –(1 + tanx)

1 + 2tanx + tanx– 1 – tanx

tanx + tanx

= RHS

LHS = RHS

Hence proved.

Question 2 :

sinx + cosx= 1 – 3 sinx cosx

Answer 2 :

Let us consider LHS: sinx+ cosx

(sinx) 3 +(cosx) 3

By using the formula, a3 + b3 =(a + b) (a2 + b2 – ab)

(sinx + cosx)[(sinx) 2 +(cosx) 2 –sinx cosx]

By using the formula, sinx + cosx= 1 and a2 + b2 = (a + b) 2 –2ab

1 × [(sinx + cosx) 2 – 2sinxcosx – sinx cosx

12 –3sinx cosx

1 – 3sinx cosx

= 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.

Question 7 :

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.

Question 8 : (sec x sec y + tan x tan y)2 – (sec x tan y + tan x sec y)2 = 1

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)] [secx sec2 y +tanx tan2 y + 2 (sec x sec y) (tan x tan y)]– [secx tan2 y + tanx sec2 y+ 2 (secx tan2 y) (tan x sec y)]

secx sec2 y – secxtan2 y + tanx tan2 y – tanxsec2 y

secx (sec2 y – tan2 y)+ tanx (tan2 y – sec2 y)

secx (sec2 y – tan2 y)– tanx (sec2 y – tan2 y)

We know, secx – tanx= 1.

secx × 1 – tanx ×1

secx – tanx

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.


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RD Chapter 5- Trigonometric Functions Ex-5.1 Contributors

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