Basic Trig ratio problems are very important when dealing with triangles. In the below questions we will learn how to find the values of the other ratio where one ratio is given.

**1.** If sec θ = 17/8 and θ is a positive acute angle, find the value of csc θ using Pythagoras theorem.

**Solution:**

Draw a right-angled ∆ ABC such that ∠ABC = θ,

Hypotenuse = BA = 17, and Adjacent side (or base) = BC = 8.Then we get,

sec θ = 17/8

Now, from the right-angled ∆ ABC we get,

AC

⇒ AC

⇒ AC

⇒ AC

⇒ AC

Therefore, AC = 15 (Since θ is a positive acute angle so, AC is also positive)

Therefore, csc θ = BA/AC

⇒ csc θ = 17/15

In this question on Trig ratio problems we will learn how to find the value of sin θ when θ is a positive acute angle.

**2.** If
tan θ +
sec θ =
2/√3 and θ
is a positive acute angle, find the value of sin θ.

**Solution: **

Given, tan θ + sec θ = 2/√3,

⇒ sin θ/cos θ + 1/cos θ = 2/√3,( Since tan θ = sin θ/cos θ and sec θ = 1/cos θ)

⇒ (sin θ + 1)/cos θ = 2/√3

⇒ √3 (sin θ + 1) = 2 cos θ

⇒ 3(sin θ + 1)⇒ 3(sin

⇒ 3 sin

⇒ 3 sin

⇒ 7 sin

⇒ 7 sin

⇒ 7 sin θ (sin θ + 1) - 1 (sin θ + 1) =0

⇒ (7 sin θ - 1)(sin θ + 1) = 0

Therefore, Either, 7 sin θ - 1 = 0
⇒ 7 sin θ = 1 ⇒ sin θ = 1/7 |
or, sin θ + 1 = 0 ⇒ sin θ = - 1 |

According to the problem, θ is a positive acute angle; so, we neglect, sin θ = -1.

Therefore, sin θ = 1/7

The below solved Trig ratio problems will help us to find the values of the ratio using trigonometric identity.

**3.**
If θ is
a positive acute angle and sec θ = 25/7, find
the value of csc θ using trigonometric identity.

**Solution: **

Given, sec θ = 25/7

Therefore, cos θ = 1/sec θ

⇒ cos θ = 1/(25/7)

⇒
cos θ =
7/25

⇒ sin

⇒ sin

⇒ sin

⇒ sin

⇒ sin

Now, taking square root on both the sides we get,

⇒ sin θ = 24/25 (Since θ is a positive acute angle so, sin θ is also positive)

Therefore, csc θ = 1/sin θ

⇒ csc θ = 1/(24/25)

⇒ csc
θ =
25/24 .

In this question on Trig ratio problems we will learn how to find the minimum value of the given T-ratio.

cos

= (cos θ)

= (cos θ - sec θ)

= (cos θ - sec θ)

(cos θ - sec θ)

Therefore, (cos θ - sec θ)

i.e., cos

Therefore, the minimum value of cos

**●** **Trigonometric Functions**

**Basic Trigonometric Ratios and Their Names****Restrictions of Trigonometrical Ratios****Reciprocal Relations of Trigonometric Ratios****Quotient Relations of Trigonometric Ratios****Limit of Trigonometric Ratios****Trigonometrical Identity****Problems on Trigonometric Identities****Elimination of Trigonometric Ratios****Eliminate Theta between the equations****Problems on Eliminate Theta****Trig Ratio Problems****Proving Trigonometric Ratios****Trig Ratios Proving Problems****Verify Trigonometric Identities****Trigonometrical Ratios of 0°****Trigonometrical Ratios of 30°****Trigonometrical Ratios of 45°****Trigonometrical Ratios of 60°****Trigonometrical Ratios of 90°****Trigonometrical Ratios Table****Problems on Trigonometric Ratio of Standard Angle****Trigonometrical Ratios of Complementary Angles****Rules of Trigonometric Signs****Signs of Trigonometrical Ratios****All Sin Tan Cos Rule****Trigonometrical Ratios of (- θ)****Trigonometrical Ratios of (90° + θ)****Trigonometrical Ratios of (90° - θ)****Trigonometrical Ratios of (180° + θ)****Trigonometrical Ratios of (180° - θ)****Trigonometrical Ratios of (270° + θ)****Trigonometrical Ratios of (270° - θ)****Trigonometrical Ratios of (360° + θ)****Trigonometrical Ratios of (360° - θ)****Trigonometrical Ratios of any Angle****Trigonometrical Ratios of some Particular Angles****Trigonometric Ratios of an Angle****Trigonometric Functions of any Angles****Problems on Trigonometric Ratios of an Angle****Problems on Signs of Trigonometrical Ratios**

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