# Prove that cos Î¸/(1 + sin Î¸) = (1 + cos Î¸ – sin Î¸)/(1 + cos Î¸ + sin Î¸)

Trigonometry is a discipline of mathematics that studies the relationships between the lengths of the sides and angles of a right-angled triangle. Trigonometric functions, also known as goniometric functions, angle functions, or circular functions, are functions that establish the relationship between an angle to the ratio of two of the sides of a right-angled triangle. The six main trigonometric functions are sine, cosine, tangent, cotangent, secant, or cosecant.

Angles defined by the ratios of trigonometric functions are known as trigonometry angles. Trigonometric angles represent trigonometric functions. The value of the angle can be anywhere between 0-360Â°.

As given in the above figure in a right-angled triangle:

**Hypotenuse:**The side opposite to the right angle is the hypotenuse, It is the longest side in a right-angled triangle and opposite to the 90Â° 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.

**Trigonometric Functions**

Trigonometry has 6 basic trigonometric functions, they are sine, cosine, tangent, cosecant, secant, and cotangent. Now letâ€™s look into the trigonometric functions. The six trigonometric functions are as follows,

sine:It is defined as the ratio of perpendicular and hypotenuse and It is represented as sin Î¸

cosine:It is defined as the ratio of base and hypotenuse and it is represented as cos Î¸

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 Î¸

cosecant:It is the reciprocal of sin Î¸ and is represented as cosec Î¸.

secant:It is the reciprocal of cos Î¸ and is represented as sec Î¸.

cotangent:It is the reciprocal of tan Î¸ and is represented as cot Î¸.>Trigonometric RatiosSin Î¸ = Perpendicular / Hypotenuse = AB/AC

Cosine Î¸ = Base / Hypotenuse = BC/AC

Tangent Î¸ = Perpendicular / Base = AB/BC

Cosecant Î¸ = Hypotenuse / Perpendicular = AC/AB

Secant Î¸ = Hypotenuse / Base = AC/BC

Cotangent Î¸ = Base / Perpendicular = BC/AB

Reciprocal IdentitiesSin Î¸ = 1/ Cosec Î¸ OR Cosec Î¸ = 1/Sin Î¸

Cos Î¸ = 1/ Sec Î¸ OR Sec Î¸ = 1/Cos Î¸

Cot Î¸ = 1/Tan Î¸ OR Tan Î¸ = 1/Cot Î¸Cot Î¸ = Cos Î¸/Sin Î¸ OR Tan Î¸ = Sin Î¸/Cos Î¸

Tan Î¸.Cot Î¸ = 1

Trigonometric Identities of Complementary and Supplementary Angles

Complementary Angles:Pair of angles whose sum is equal to 90Â°Supplementary Angles:Pair of angles whose sum is equal to 180Â°

Identities of Complementary angles aresin (90Â° – Î¸) = cos Î¸

cos (90Â° – Î¸) = sin Î¸

tan (90Â° – Î¸) = cot Î¸

cot (90Â° – Î¸) = tan Î¸

sec (90Â° – Î¸) = cosec

p>cosec (90Â° – Î¸) = sec Î¸

Identities of supplementary anglessin (180Â° – Î¸) = sin Î¸

cos (180Â° – Î¸) = – cos Î¸

tan (180Â° – Î¸) = – tan Î¸

cot (180Â° – Î¸) = – cot Î¸

sec (180Â° – Î¸) = – sec Î¸

cosec (180Â° – Î¸) = – cosec Î¸

Prove that cosÎ¸/(1 + sin Î¸) = (1 + cos Î¸ – sin Î¸)/(1 + cos Î¸ + sin Î¸)

Solution:We have cos Î¸/(1 + sin Î¸) = (1 + cos Î¸ – sin Î¸)/(1 + cos Î¸ + sin Î¸)

Fi

we will take LHS = cos Î¸/(1 + sin Î¸)= {[cos Î¸/(1 + sin Î¸)] Ã— [(1-sin Î¸)/(1-sin Î¸)]}

= [cos Î¸(1-sin Î¸)]/(1- sin

^{2}Î¸)= [cos Î¸(1-sin Î¸)]/ cos

^{ 2}Î¸{ 1- sin^{2}Î¸ = cos^{2}Î¸ }= (1-sin Î¸) / cos Î¸

= (1/ cos Î¸) – (sin Î¸/cos Î¸)

LHS = sec Î¸ – tan Î¸Now

RHS = (1 + cos Î¸ – sin Î¸)/(1 + cos Î¸ + sin Î¸)

= [(1 + cos Î¸ – sin Î¸)/(1 + cos Î¸ + sin Î¸) Ã— (1 + cos Î¸ – sin Î¸)/(1 + cos Î¸ – sin Î¸)]

= [(1 + cos Î¸ – sin Î¸)

¸)^{2}] /[(1 + c^{2}+ (sin Î¸)^{2}]= [(1 + cos Î¸)

^{2}+ sin^{2}Î¸ – 2 (1 + cos Î¸)(sin Î¸) ]/(1 + cos^{ 2}Î¸ + 2cos Î¸ – sin^{2}Î¸)= [(1 + cos

^{2}Î¸ + sin^{2}Î¸ + 2 cos Î¸ – 2 (1+cos Î¸)(sin Î¸)]/[1+ cos^{2}Î¸ + 2cos Î¸ -(1-cos^{2}Î¸)]{… 1- cos^{2}Î¸ = sin^{2}Î¸}= [2 +2 cos Î¸ – 2 (sin Î¸) (1 + cos Î¸) ]/[1 +cos

^{2}Î¸ + 2cos Î¸ – 1 +cos^{2}Î¸]{sin^{2 }Î¸ + cos^{2}Î¸ = 1}= [2 + 2cos Î¸ – 2sin Î¸ (1 + cos Î¸)]/[2cos

^{2}Î¸ + 2cos Î¸]= [2 + 2cos Î¸ – 2sin Î¸ (1 + cos Î¸)] / [2 cos Î¸ (cos Î¸ +1)]

= [2 (1 + cos Î¸) – 2 sin Î¸(1 + cos Î¸)] / [2 cos Î¸ (cos Î¸ +1)]

= [2 (1 + cos Î¸) (1 – sin Î¸)]/[2 cos Î¸ (cos Î¸ +1)]

= (1 – sin Î¸) / cos Î¸

= (1/ cos Î¸) – (sin Î¸/ cos Î¸)

~~g>RHS = sec Î¸ – tan Î¸~~

Therefore LHS = RHS

cos Î¸/(1 + sin Î¸) = (1 + cos Î¸ – sin Î¸)/(1 + cos Î¸ + sin Î¸)

Hence proved

Similar Questions

Question 1: If x sin^{3}Î¸ + y cos^{3}Î¸ = sin Î¸ cos Î¸ and x sin Î¸ â€“ y cos Î¸ = 0, then prove that x^{2}+ y^{2}= 1, (where, sin Î¸ â‰ 0 and cos Î¸ â‰ 0).

Solution:here we have, x sin

^{3}Î¸ + y cos^{3}Î¸ = sin Î¸ cos Î¸Given : x sin

^{3}Î¸ + y cos^{3}Î¸ = sin Î¸ cos Î¸â‡’ (x sin Î¸) sin

^{2}Î¸ + (y cos Î¸) cos^{2}Î¸ = sin Î¸ cos Î¸â‡’ (x sin Î¸) sin

^{2}Î¸ + (x sin Î¸) cos^{2}Î¸ = sin Î¸ cos Î¸ (âˆµ y cos Î¸ = x sin Î¸)â‡’ x sin Î¸(sin

p>2 Î¸ + cos^{2}Î¸ + cos^{2}Î¸) = sin Î¸ cos Î¸ (si^{2}Î¸ = 1)â‡’ x sin Î¸ = sin Î¸ cos Î¸

â‡’

x = cos Î¸â€¦.(eq. 1)now another trigono eq we have x sin Î¸ â€“ y cos Î¸ = 0

we can write it as x sin Î¸ = y cos Î¸

from eq. 1 we have x = cos Î¸ so put in above eq. x sin Î¸ = y cos Î¸

so x sin Î¸ = y cos Î¸

cos Î¸ sin Î¸ = y cos Î¸

y = sin Î¸ eq. 2now by squaring and adding both the equation 1 & 2

x = cos Î¸ & y = sin Î¸

x

^{2}= cos^{2}Î¸ & y^{2}= sin^{2}Î¸so now x

^{2 }+ y^{2}= cos^{2}Î¸ + sin^{2}Î¸{ cos^{2 }Î¸ + sin^{2}Î¸ = 1 }

x^{2}+ y^{2}= 1

Hence proved

Question 2: Evaluate (Sin 30Â° – Sin 90Â° + 2 Cos 0Â°) / Tan 30Â° Tan 60Â°?

Solution:Here we have (Sin 45Â° – Sin 90Â° + 2 Cos 0Â°) / Tan 45Â° Tan 60Â°

As per the trigonometric values

(Sin 45Â° – Sin 90Â° +2 Cos 0Â°) / Tan 45Â° Tan 60Â°

= (1/âˆš2 – 1 + 2 Ã— 1) / 1 Ã— âˆš3

= (1/âˆš2 – 1 + 2) / âˆš3

= (1/âˆš2 + 1) / âˆš3

= (1+âˆš2 / âˆš2) / âˆš3