In this page trigonometry problems set4 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 14
Prove that sin θ (1 + tan θ) + cos θ (1 + cot θ) = sec θ + cosec θ
Solution:
L.H.S
= sin θ (1 + tan θ) + cos θ (1 + cot θ)
To simplify this first we are going to apply the formulas for tanθ and cotθ
Instead of tan θ we can write sin θ/cos θ.
Like that instead of cot θ we can write cos θ/sin θ.
= sin θ (1 + sin θ/cos θ) + cos θ (1 + cos θ/sin θ)
Since the denominators are not same let us take L.C.M for both fractions.
Now we are going to take (cos θ + sin θ) from the numerator.
Now we are plugging the value of cos
² θ + sin
² θ that is 1.
both cos θ and sin θ will get cancelled in the numerator and denominators from the first and second fraction respectively.
We have used reciprocal formula ti get the value of 1/sin θ and 1/cos θ.
Question 15
Prove that (1 - sin² θ) sec² θ = 1
Solution:
L.H.S
= (1 - sin² θ) sec² θ
we can write cos² θ instead of 1 - sin² θ form the identity cos² θ + sin² θ = 1.
= (cos² θ) sec² θ
in the next step we are going to write 1/cos²θ instead of sec² θ using reciprocal formula.
= (cos² θ) (1/cos² θ)
= (cos² θ/cos² θ)
= 1
R.H.S
Both cos² θ both are in the numerator and denominator will get cancelled. Finally we get 1.
Question 16
Prove that (1 + tan² θ) cos² θ = 1
Solution:
L.H.S
= (1 + tan² θ) cos² θ
From this identity we come to know the value of (1 + tan² θ) that is sec²θ.
= (sec² θ) cos² θ trigonometry problems set4
= (1/cos² θ) cos² θ
= (cos² θ/cos² θ)
= 1
R.H.S