Solved Examples
Question 1: Prove the following identity:
Solution:
Solution:
LHS = (sec - cose)(1 + tan + cot)
= ()()
[Using identities, sec = , cose = , tan = and cot = ]
=
=
[Using, = 1]
= tan sec - cotcosc
= RHS
= ()()
[Using identities, sec = , cose = , tan = and cot = ]
=
=
[Using, = 1]
= tan sec - cotcosc
= RHS
Question 2:Prove the following identity:
Solution:
Solution:
LHS = 1 + 2sec2A tan2A - sec4A - tan4A
= 1 - (sec4A - 2sec2A tan2A + tan4A)
[Using identity, sec2A - tan2A = 1]
= 1 - 1
= 0
= RHS
= 1 - (sec4A - 2sec2A tan2A + tan4A)
[Using identity, sec2A - tan2A = 1]
= 1 - 1
= 0
= RHS
Question 3: Prove the following identity:
Solution:
Solution:
LHS = [][]
= [][]
[Using identities, sec = , cose = , sin2A - cos2A = (sin A + cos A)(sin A - cos A) and sin3A + cos3A = (sin A + cos A)(sin2 A - sin A cos A + cos2A)]
= [][] [By cancelling common terms]
= sin A[] * []
[Using identity sin2 A + cos2 A = 1]
= sin A (By cancelling common terms)
= RHS
= [][]
[Using identities, sec = , cose = , sin2A - cos2A = (sin A + cos A)(sin A - cos A) and sin3A + cos3A = (sin A + cos A)(sin2 A - sin A cos A + cos2A)]
= [][] [By cancelling common terms]
= sin A[] * []
[Using identity sin2 A + cos2 A = 1]
= sin A (By cancelling common terms)
= RHS
Trigonometry Test Questions
Back to TopSolved Examples
Question 1: Prove the following identity:
Solution:
Solution:
LHS = (sec - 1)2 - (tan - sin)2
= ( - 1)2 - ( - sin)2
= ()2 - (1 - sin * )2
= ()2 - (1 - cos)2
= (1 - cos)2 []
= ( - 1)2 - ( - sin)2
= ()2 - (1 - sin * )2
= ()2 - (1 - cos)2
= (1 - cos)2 []
Question 2:Prove the following identity:
Solution:
Solution:
LHS = +
= +
= +
Trigonometry Sample Questions
Back to TopSolved Examples
Question 1:
Solution:
Solution:
If tan A + sin A = m .................. (1)
tan A - sin A = n ................... (2)
Step 1:Adding (1) and (2)
Step 2:
Subtracting (2) from (1)
Now
[Using, (a + b)2 = a2 + b2 + 2ab and (a - b)2 = a2 + b2 - 2ab]
tan A - sin A = n ................... (2)
Step 1:Adding (1) and (2)
Step 2:
Subtracting (2) from (1)
Now
[Using, (a + b)2 = a2 + b2 + 2ab and (a - b)2 = a2 + b2 - 2ab]
Question 2:
Solution:
= * cos
= ( - 1)cos
Solution:
= * cos
= ( - 1)cos
Question 3:If x sin3 + y cos3 = sin cos and x sin - y cos = 0. Prove that x2 + y2 = 1.
Solution:
Solution:
x sin3 + y cos3 = sin cos ..................... (i)
x sin - y cos = 0 ...................... (ii)
from (i) and (iii)
y = sin .................. (iv)
From (iii) and (iv), x = cos
=> x2 + y2 = sin2 + cos2
x sin - y cos = 0 ...................... (ii)
from (i) and (iii)
y = sin .................. (iv)
From (iii) and (iv), x = cos
=> x2 + y2 = sin2 + cos2