BASIC TRIGONOMETRIC IDENTITIES WORKSHEET

Problem 1 : 

If p = sinθ + cosθ, q = sinθ - cosθ, then find the value of

(p2 + q2)

Problem 2 : 

If x = secθ + tanθ, q = secθ + tan θ, then find the value of

xy

Problem 3 : 

If a = cscθ + cotθ, b = cscθ - cotθ, then find the value of

(a2 - b2)

Problem 4 : 

Prove that 

Problem 5 : 

Prove that 

(secA - cosecA)(1 + tanA + cotA)  =  tanAsecA - cotA cosecA

Problem 6 : 

Eliminate θ from

acosθ  =  b

csinθ  =  d

where a, b, c and d are constants.

Answers

Problem 1 : 

If p = sinθ + cosθ, q = sinθ - cosθ, then find the value of

(p2 + q2)

Solution :

p2 + q=  (sinθ + cosθ)2 + (sinθ - cosθ)2

p2 + q2  =  (sin2θ + cos2θ + 2sinθcosθ) + (sin2θ + cos2θ - 2sinθcosθ)

p2 + q2  =  sin2θ + cos2θ + 2sinθcosθ + sin2θ + cos2θ - 2sinθcosθ

p2 + q2  =  2sin2θ + 2cos2θ

p2 + q2  =  2(sin2θ + cos2θ)

p2 + q2  =  2(1)

p2 + q2  =  2

Problem 2 : 

If x = secθ + tanθ, q = secθ + tan θ, then find the value of 

xy

Solution :

xy  =  (secθ + tanθ)(secθ - tanθ)

xy  =  sec2θ - tan2θ

xy  =  1

Problem 3 : 

If a = cscθ + cotθ, b = cscθ - cotθ, then find the value of

(a2 - b2)

Solution :

a2 - b2  =  (cscθ + cotθ)2 - (cscθ - cotθ)2

a2 + b2  =  (csc2θ + cot2θ + 2cscθcotθ) - (csc2θ + cot2θ - 2cscθcotθ)

a2 + b2  =  csc2θ + cot2θ + 2cscθcotθ - csc2θ - cot2θ + 2cscθcotθ

a2 - b2  =  4cscθcotθ

Problem 4 : 

Prove that 

Solution :

Problem 5 : 

Prove that 

(secA - cosecA)(1 + tanA + cotA)  =  tanAsecA - cotA cosecA

Solution :

From (i) and (ii), we get the required result.

Problem 6 : 

Eliminate θ from

acosθ  =  b

csinθ  =  d

where a, b, c and d are constants.

Solution :

acosθ  =  b

Multiply each side by c. 

accosθ  =  bc

Square each side. 

(accosθ)2  =  (bc)2

a2c2cos2θ  =  b2c2 -----(1)

csinθ  =  d

Multiply each side by a. 

acsinθ  =  ad

Square each side. 

(acsinθ)2  =  (ad)2

a2c2sin2θ  =  a2d-----(2)

Add (1) and (2). 

a2c2cos2θ + a2c2 sin2θ  =  b2c+  a2d2

a2c2(cos2θ + sin2θ)  =  b2c2  +  a2d2

a2c2(1)  =  b2c2  +  a2d2

a2c2  =  b2c2  +  a2d2

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