In this page trigonometry problems set3 we are going to see practice questions in this topic.Here we you can find solution with detailed explanation.
Let us see trigonometric-identities
These identities are applied in both ways ,left to right or right to left.So we have to memories all the identities.
Question 10
Prove that (sec θ + cos θ) (sec θ - cos θ) = tan²θ + sin²θ
Solution:
L.H.S
= (sec θ + cos θ) (sec θ - cos θ)
Let us compare this terms with the following algebraic identity.Here instead of a we have sec θ and instead of b we have cos θ.
= (sec² θ - cos² θ)
trigonometry problems set3 | |
From this identity we can derive formula for sec² θ that is 1 + tan² θ and we can derive for cos² θ that is 1 - sin² θ.
= [(1 + tan² θ) - (1 - sin² θ)]
distributing negative for 1 - sin² θ we are getting
= [1 + tan² θ - 1 + sin² θ]
= tan² θ + sin² θ
R.H.S
Question 11
Simplify the following
Solution:
Since the denominators are not same we need to take L.C.M for two fractions.
removing bracket we get
negative sin θ and positive sin θ will get cancelled.
1 - sin² θ will become cos² θ.
Question 12
Simplify the following
Solution:
Since the denominators are not same we need to take L.C.M for two fractions.
applying the formula (a+b)² to expand (1+sin θ)². Here instead of a we have 1 and instead of b we have sin θ.
now apply the value for sin²θ + cos²θ that is 1.
Now we can take 2 from both terms in the numerator. Then we will get
both (1+sinθ) which is in numerator and denominator will get cancelled.
Question 13
Prove the following
Solution:
we can use 1- cos²θ instead of sin²θ.
comparing 1 - cos²θ with algebraic formula a² - b² = (a + b)(a - b) we can write (1 + cosθ) (1 + cosθ)
both (1 + cosθ) which are in the numerator and denominator will get cancelled.