To learn trigonometric identities in detail,
Problem 1 :
Prove :
(1 - cos2θ) csc2θ = 1
Solution :
Let A = (1 - cos2θ) csc2θ and B = 1.
A = (1 - cos2θ) csc2θ
Because sin2θ + cos2θ = 1, we have
sin2θ = 1 - cos2θ
Then,
A = sin2θ ⋅ csc2θ
A = sin2θ ⋅ (1/sin2θ)
A = sin2θ /sin2θ
A = 1
A = B (Proved)
Problem 2 :
Prove :
sec θ √(1 - sin2θ) = 1
Solution :
Let A = sec θ √(1 - sin2θ) and B = 1.
A = sec θ √(1 - sin2θ)
Because sin2θ + cos2θ = 1, we have
cos2θ = 1 - sin2θ
Then,
A = sec θ √cos2θ
A = sec θ ⋅ cos θ
A = sec θ ⋅ (1/sec θ)
A = sec θ / sec θ
A = 1
A = B (Proved)
Problem 3 :
Prove :
tan θ sin θ + cos θ = sec θ
Solution :
Let A = tan θ sin θ + cos θ and B = sec θ.
A = tan θ sin θ + cos θ
A = (sin θ/cos θ) ⋅ sin θ + cos θ
A = (sin2θ/cos θ) + cos θ
A = (sin2θ/cos θ) + (cos2θ/cosθ)
A = (sin2θ + cos2θ) / cos θ
A = 1 / cos θ
A = sec θ
A = B (Proved)
Problem 4 :
Prove :
(1 - cos θ)(1 + cos θ)(1 + cot2θ) = 1
Solution :
Let A = (1 - cos θ)(1 + cos θ)(1 + cot2θ) = 1 and B = 1.
A = (1 - cos θ)(1 + cos θ)(1 + cot2θ)
A = (1 - cos2θ)(1 + cot2θ)
Because sin2θ + cos2θ = 1, we have
sin2θ = 1 - cos2θ
Then,
A = sin2θ ⋅ (1 + cot2θ)
A = sin2θ + sin2θ ⋅ cot2θ
A = sin2θ + sin2θ ⋅ (cos2θ/sin2θ)
A = sin2θ + cos2θ
A = 1
A = B (Proved)
Problem 5 :
Prove :
cot θ + tan θ = sec θ csc θ
Solution :
Let A = cot θ + tan θ and B = sec θ csc θ.
A = cot θ + tan θ
A = (cos θ/sin θ) + (sin θ/cos θ)
A = (cos2θ/sin θ cos θ) + (sin2θ/sin θ cos θ)
A = (cos2θ + sin2θ) / sin θ cos θ
A = 1 / sin θ cos θ
A = (1/cos θ) ⋅ (1/sin θ)
A = sec θ csc θ
A = B (Proved)
Problem 6 :
Prove :
cos θ/(1 - tan θ) + sin θ/(1 - cot θ) = sin θ + cos θ
Solution :
Let A = cos θ/(1 - tan θ) + sin θ/(1 - cot θ) and
B = sin θ + cos θ
A = cos θ/{1 - (sin θ/cos θ)} + sin θ/{1 - (cos θ/sin θ)}
A = cos2θ/(cos θ - sin θ) + sin2θ/(sin θ - cos θ)
A = cos2θ/(cos θ - sin θ) - sin2θ/(cos θ - sin θ)
A = (cos2θ - sin2θ) / (cos θ - sin θ)
A = [(cos θ + sin θ)(cos θ - sin θ)] / (cos θ - sin θ)
A = (cos θ + sin θ)
A = B (Proved)
Problem 7 :
Prove :
tan4θ + tan2θ = sec4θ - sec2θ
Solution :
Let A = tan4θ + tan2θ and B = sec4θ + sec2θ.
A = tan4θ + tan2θ
A = tan2θ (tan2θ + 1)
We know that,
tan2θ = sec2θ - 1
tan2θ + 1 = sec2θ
Then,
A = (sec2θ - 1)(sec2θ)
A = sec4θ - sec2θ
A = B (Proved)
Problem 8 :
Prove :
√{(sec θ – 1)/(sec θ + 1)} = cosec θ - cot θ
Solution :
Let A = √{(sec θ – 1)/(sec θ + 1)} and B = cosec θ - cot θ.
A = √{(sec θ – 1)/(sec θ + 1)}
A = √[{(sec θ - 1) (sec θ - 1)}/{(sec θ + 1) (sec θ - 1)}]
A = √{(sec θ - 1)2 / (sec2θ - 1)}
A = √{(sec θ - 1)2 / tan2θ}
A = (sec θ – 1)/tan θ
A = (sec θ/tan θ) – (1/tan θ)
A = {(1/cos θ)/(sin θ/cos θ)} - cot θ
A = {(1/cos θ) ⋅ (cos θ/sin θ)} - cot θ
A = (1/sin θ) - cot θ
A = cosec θ - cot θ
A = B (Proved)
Problem 9 :
Prove :
(1 - sin A)/(1 + sin A) = (sec A - tan A)2
Solution :
Let A = (1 - sin A)/(1 + sin A) and B = (sec A - tan A)2.
A = (1 - sin A) / (1 + sin A)
A = (1 - sin A)2 / (1 - sin A) (1 + sin A)
A = (1 - sin A)2 / (1 - sin2A)
A = (1 - sin A)2 / (cos2A)
A = (1 - sin A)2 / (cos A)2
A = {(1 - sin A) / cos A}2
A = {(1/cos A) - (sin A/cos A)}2
A = (sec A – tan A)2
A = B (Proved)
Problem 10 :
Prove :
(tan θ + sec θ - 1)/(tan θ - sec θ + 1) = (1 + sin θ)/cos θ
Solution :
Let A = (tan θ + sec θ - 1)/(tan θ - sec θ + 1) and
B = (1 + sin θ)/cos θ.
A = (tan θ + sec θ - 1)/(tan θ - sec θ + 1)
A = [(tan θ + sec θ) - (sec2θ - tan2θ)]/(tan θ - sec θ + 1)
A = {(tan θ + sec θ) (1 - sec θ + tan θ)}/(tan θ - sec θ + 1)
A = {(tan θ + sec θ) (tan θ - sec θ + 1)}/(tan θ - sec θ + 1)
A = tan θ + sec θ
A = (sin θ/cos θ) + (1/cos θ)
A = (sin θ + 1)/cos θ
A = (1 + sin θ)/cos θ
A = B, (Proved)
Apart from the stuff given in this section, if you need any other stuff in math, please use our google custom search here.
Kindly mail your feedback to v4formath@gmail.com
We always appreciate your feedback.
©All rights reserved. onlinemath4all.com
Jul 02, 22 07:59 AM
Horizontal Expansions and Compressions - Concept - Examples
Jul 02, 22 07:41 AM
Reflection Through Y Axis - Concept - Examples with Step by Step Explanation
Jul 02, 22 07:34 AM
Reflection Through X Axis - Concept - Example with Step by Step Explanation