Trigonometry Problems set3

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.





Identities of Trigonometry

Let us see trigonometric-identities

  1. sin² θ  + cos² θ = 1
  2. sin² θ  = 1 - cos² θ
  3. cos² θ = 1 - sin² θ
  4. Sec² θ - tan² θ = 1
  5. Sec² θ  = 1 +  tan² θ
  6. tan² θ  =  Sec² θ - 1
  7. Cosec² θ - cot² θ = 1
  8. Cosec² θ = 1 + cot² θ
  9. cot² θ =  Cosec² θ - 1

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.








Trigonometry Problems Set3 to Trigonometry