Trigonometry Problems set2

In this page trigonometry problems set2 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 6

Prove that cosec θ √(1 - cos²θ) = 1

Solution:

L.H.S

                       = cosec θ √(1 - cos²θ)

We can write sin ² θ instead of 1 - cos²θ.

                       =  cosec θ √sin ² θ 

We can take one sin θ from the radical sign.

                       =  cosec θ sin θ

we can write 1/sin θ instead of cosec θ.

                       =  (1/sin θ) sin θ

                       = (sin θ/sin θ)

                       = 1

                          R.H.S


Question 7

Prove that (1 - cos²θ) sec²θ = tan²θ

Solution:

L.H.S

                       = (1 - cos²θ) sec²θ

We can write sin ² θ instead of 1 - cos²θ.

                       =  sin²θ sec²θ

trigonometry problems set2

we can write 1/cos ²θ instead of sec² θ.

                       =  sin²θ (1/cos²θ)

                       =  (sin²θ /cos²θ)

                       = tan²θ

                          R.H.S


Question 8

Prove that (sec²θ-1) (cosec²θ-1) = 1

Solution:

L.H.S

                       = (sec²θ-1) (cosec²θ-1)

we can write tan²θ instead of (sec² θ - 1).

we can write cot²θ instead of (cosec² θ - 1).

                       = (tan²θ) (cot²θ)        

we can write 1/tan²θ instead of cot² θ.

                       = (tan²θ) (1/tan²θ)

                       = (tan²θ/tan²θ)

                       = 1

                          R.H.S


Question 9

Prove that secθ (1 - sinθ) (secθ + tan θ) = 1

Solution:

L.H.S

                       = sec θ (1 - sin θ) (secθ + tan θ)

We can write 1/cos θ instead of sec θ

                      = (1/cos θ) (1 - sin θ) (secθ + tan θ)

We can write sin θ/cos θ instead of tan θ

                      = (1/cos θ) (1 - sin θ) [(1/cos θ) + (sin θ/cosθ)]

Now we are going to take L.C.M

                      = (1/cos θ) (1 - sin θ) [(1 + sin θ)/cosθ]

                      = (1 - sin θ)/cos θ  [(1 + sin θ)/cosθ]

From this identity we can write 1 - sin²θ instead of (1 + sinθ) (1 - sinθ)

                      = (1 - sin²θ)/cos² θ 

We can write cos² θ  instead of 1 - sin² θ

                      = (cos² θ )/cos² θ 

                      = 1 

                          R.H.S








Trigonometry Problems Set2 to Trigonometry