Cosec 150.
Trigonometry is a subject of mathematics that studies the relationships between the lengths of the sides and angles of a rightangled triangle. Trigonometric functions, as well known every bit goniometric functions, bending functions, or circular functions, are functions that constitute the relationship between an bending to the ratio of two of the sides of a correctangled triangle. The six primary trigonometric functions are sine, cosine, tangent, cotangent, secant, or cosecant.
Angles defined past the ratios of trigonometric functions are known every bit trigonometry angles. Trigonometric angles represent trigonometric functions. The value of the angle can be anywhere between 0360°.
Equally given in the above figure in a rightangled triangle:

Hypotenuse:
The side opposite to the correct angle is the hypotenuse, It is the longest side in a rightangled triangle and opposite to the xc° angle. 
Base:
The side on which angle C lies is known as the base. 
Perpendicular:
It is the side opposite to angle C in consideration.
Daftar Isi:
Trigonometric Functions
Trigonometry has vi basic trigonometric functions, they are sine, cosine, tangent, cosecant, secant, and cotangent. Now let’south look into the trigonometric functions. The halfdozen trigonometric functions are equally follows,

sine:
It is divers as the ratio of perpendicular and hypotenuse and It is represented as sin θ 
cosine:
It is defined as the ratio of base of operations and hypotenuse and it is represented equally cos θ 
tangent:
It is defined as the ratio of sine and cosine of an angle. Thus the definition of tangent comes out to be the ratio of perpendicular and base and is represented as tan θ 
cosecant:
It is the reciprocal of sin θ and is represented every bit cosec θ. 
secant:
It is the reciprocal of cos θ and is represented as sec θ. 
cotangent:
It is the reciprocal of tan θ and is represented as cot θ.
According to above given paradigm, Trigonometric Ratios are
Sin θ
= Perpendicular / Hypotenuse = AB/AcCosine θ
= Base of operations / Hypotenuse = BC / AirconditioningTangent θ
= Perpendicular / Base = AB / BCCosecant θ
= Hypotenuse / Perpendicular = AC/ABSecant θ
= Hypotenuse / Base = Ac/BCCotangent θ
= Base / Perpendicular = BC/AB
Reciprocal Identities
Sin θ = i/ Cosec θ Or Cosec θ = ane/ Sin θ
Cos θ = i/ Sec θ Or Sec θ = 1 / Cos θ
Cot θ = 1 / Tan θ Or Tan θ = 1 / Cot θ
Cot θ = Cos θ / Sin θ Or Tan θ = Sin θ / Cos θ
Tan θ.Cot θ = 1
Values of Trigonometric Ratios
0° 
xxx° 
45° 
sixty° 
90° 

Sin θ 
1/two 
1/√ii 
√3/2 
1 

Cos θ 
one 
√three/2 
1/√ii 
1/2 

Tan θ 
1/√3 
1 
√3 
— 

Cosec θ 
— 
two 
√2 
2/√3 
1 
Sec θ 
1 
2/√3 
√2 
2 
— 
Cot θ 
— 
√3 
1 
ane/√3 
Trigonometric Identities of Complementary and Supplementary Angles

Complementary Angles:
Pair of angles whose sum is equal to 90° 
Supplementary Angles:
Pair of angles whose sum is equal to 180°
Identities of Complementary angles are
sin (90° – θ) = cos θ
cos (ninety° – θ) = sin θ
tan (90° – θ) = cot θ
cot (xc° – θ) = tan θ
sec (90° – θ) = cosec θ
cosec (ninety° – θ) = sec θ
Identities of supplementary angles
sin (180° – θ) = sin θ
cos (180° – θ) = – cos θ
tan (180° – θ) = – tan θ
cot (180° – θ) = – cot θ
sec (180° – θ) = – sec θ
cosec (180° – θ) = – cosec θ
Quadrants of trigonometry
Find the value of sin 135°
cosec 225°
tan 150°
cot 315°
Solution:
Hither nosotros have sin 135° cosec 225° tan150° cot315°
We can write as Sin (90+45) cosec (180+45) tan (90+60) cot (360 – 45) { as per the quadrants and trigonometric values }
= Cos 45° Cosec 45° Cot 60° (Cot 45°)
= i/√2
×
√ii
×
1/√three
×
1= – 1/√iii
The value of sin 135 cosec 225 tan150 cot315 is – 1/√three
Similar Questions
Question 1: Find the value of Sin 60°
Cos 30°
+ Cos 60°
Sin 30°?
Solution:
Here we accept Sin threescore° Cos 30° + Cos lx° Sin 30°
So Sin sixty° Cos thirty° + Cos 60° Sin thirty°
= √3/2 × √3/2 + ane/two × 1/2
= 3/4 + 1/4
= 1
Question 2: Evaluate (Sin 30°
– Sin 90°
+ two Cos 0°) / Tan 30°
Tan 60°?
Solution:
Here we have (Sin 30° – Sin 90° + ii Cos 0°) / Tan 30° Tan 60°
As per the trigonometric values
(Sin 30° – Sin 90° +two Cos 0°) / Tan 30° Tan lx°
= (1/ii – 1 + ii × 1 ) / ane/√3 × √3
= (1/2 – i + two ) / 1
= 3/2
Cosec 150
Source: https://www.geeksforgeeks.org/findthevalueofsin135cosec225tan150cot315/