Trigonometry Problems set5

In this page trigonometry problems set5 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 17

Prove that (1 + tan²θ) (1 - sin θ) (1 + sin θ) = 1

Solution:

L.H.S

                      = (1 + tan²θ) (1 - sin θ) (1 + sin θ)

Here (1 + tan²θ) can be written as sec²θ using the identity.

                      = (sec²θ) (1 - sin θ) (1 + sin θ)

Next we can compare (1 - sin θ) (1 + sin θ) with the algebraic formula (a+b) (a-b)

instead of a we have 1 and instead of b we have sin θ.

                      = (sec²θ) (1² - sin² θ)

                      = (sec²θ) (1 - sin² θ)

                      = (sec²θ) (cos² θ)

in the next step we are going to write sec²θ as 1/cos² θ using reciprocal formula.  

                      = (1/cos²θ) (cos² θ)

                      = (cos²θ/cos²θ)

                      = 1

                         R.H.S


Question 18

Prove that (1 - sin θ) ÷ (1 + sin θ) = (sec θ - tan θ)²

Solution:

L.H.S

                      = (1 - sin θ)/(1 + sin θ)

In the first step we need to multiply both numerator and denominator by the conjugate of the denominator of the given fraction. The denominator of the given fraction is (1 + sin θ) then its conjugate must be 1 - sin θ.

trigonometry problems set5

we can write 1 - sin² θ as cos² θ. Now we can take only one square instead of putting squares for both numerator and denominator. So that we will get

In the next step we have written this as two fractions.

By using the above identities we can write 

                           = (sec θ - tan θ)²

                              R.H.S  


Question 19

Prove that tan² θ + cot ² θ + 2 = sec² θ + cosec² θ

Solution:

L.H.S

                             = tan² θ + cot ² θ + 2

From the above identity we can derive the formula for tan² θ that is sec² θ - 1 and we can derive for cot² θ that is cosec² θ - 1.

                             = sec² θ - 1 + cosec² θ - 1 + 2

                             = sec² θ + cosec² θ - 1 - 1 + 2

                             = sec² θ + cosec² θ - 2 + 2   

                             = sec² θ + cosec² θ

                               R.H.S








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