# Trigonometry Problems set6

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In this page trigonometry problems set6 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

**sin² θ + cos² θ = 1****sin² θ = 1 - cos² θ****cos² θ = 1 - sin² θ****Sec² θ - tan² θ = 1****Sec² θ = 1 + tan² θ ****tan² θ = Sec² θ - 1****Cosec² θ - cot² θ = 1****Cosec² θ = 1 + cot² θ****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 20**

Simplify the following

**Solution:**

Since the denominators are not same we need to take L.C.M.

In the denominator we have (1 + sin θ) and (1 - sin θ) and it looks like the algebraic formula (a+b) (a-b) . So we have used the formula and got 1² - sin² θ. In the numerator multiplying (1 + sin θ) by cos θ and (1 - sin θ) by cos θ we get

negative cos θ sin θ and positive cos θ sin θ will get canceled.

identity for 1² - sin² θ is cos² θ

**Question 21**

Simplify the following

**Solution:**

L.H.S

Since the denominators are not same we need to take L.C.M.

In the denominator we have (cosec θ + 1) and (cosec θ - 1) and it looks like
the algebraic formula (a+b) (a-b) . So we have used the formula and
got cosec² θ - 1. In the numerator multiplying (cosec θ - 1) by cos θ and (cosec θ + 1) by cos θ we get

negative cos θ and positive cos θ will get canceled.

identity for cosec² θ - 1 is cot² θ.

cot² θ can be written as cos² θ/sin² θ

we have two fractions in both numerator and denominator .We have written the fraction in the numerator as it is and the fraction which is in denominator can be written as its reciprocal.

**Question 22**

Simplify the following

**Solution:**

= (1 - cos θ)/(1 + cos 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 + cos θ) then its conjugate must
be 1 - cos θ.

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

= (Cosec θ - cot θ)²

Trigonometry Problems Set6 to Trigonometry