# Trigonometric Equations

In this section we will discuss trigonometric equations.
An equation involving trigonometric ratios of an angle θ (say) is said to be a trigonometric equations, if it is satisfied for all values of θ for which the given trigonometric ratios are defined.
Some Trigonometric-equations (Identities) are as follows :

 1) sin2θ + cos2θ= 1 2) sin2θ = 1 - cos2θ 3) cos2θ = 1 - sin 2θ 4) 1 + tan2θ = sec2θ 5) tan2θ = sec2θ - 1 6) sec2θ - tan2θ = 1 7) 1 + cot2θ = csc2θ 8) cot2θ = csc2θ - 1 9) csc2θ - cot2θ = 1 10) sec θ + tan θ = 1 / ( sec θ - tan θ) 11) csc θ + cot θ = 1 / ( csc θ - cot θ)

These Trigonometric equations are true for any angle θ for which the trigonometric ratios are meaningful.

Some Solved Examples on Trigonometric Equations :

1) 1 / (1 + sin θ) + 1 / ( 1 – sin θ ) = 2 sec
2 θ

Solution :
Consider 1 / (1 + sin θ) + 1 / ( 1 – sin θ )
LCM = (1 + sin θ)( 1 – sin θ )
= [1 (1 – sin θ) + 1 ( 1 + sin θ )] / (1 + sin θ)( 1 – sin θ)
= ( 1 – sin θ + 1 + sin θ ) / (1 + sin θ)( 1 – sin θ )
= 2 / (1 – sin
2 θ ) [ use identity (a + b) ( a – b ) = a 2 - b 2 ]
= 2 / cos
2 θ
= 2 sec
2 θ
∴ 1 / (1 + sin θ) + 1 / ( 1 – sin θ ) = 2 sec
2 θ
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2) Prove that : cos θ / ( 1- sin θ ) = ( 1 + sin θ) / cos θ

Solution :
Consider cos θ / ( 1- sin θ )
Multiply top (numerator) and bottom(denominator) by 1 + sin θ
= cos θ( 1 + sin θ) / ( 1- sin θ )( 1 + sin θ)
= cos θ ( 1 + sin θ) / ( 1- sin
2 θ )
[ using this identity (a + b)(a – b ) = a
2 - b 2 ]
= cos θ ( 1 + sin θ) / cos
2 θ )
= (1 + sin θ) / cos θ
∴ cos θ / ( 1- sin θ ) = ( 1 + sin θ) / cos θ
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3) Cos
2 θ + 1 / ( 1 + cot 2 θ ) = 1

Solution :
Consider Cos
2 θ + 1 / ( 1 + cot 2 θ )
= Cos
2 θ + 1/ csc 2 θ [ since 1 + cot 2 θ = csc 2 θ]
= cos
2 θ + sin 2 θ [ since 1/ csc 2 θ = sin 2 θ ]
= 1 [ since sin
2 θ + cos 2 θ= 1 ]
∴ Cos
2 θ + 1 / ( 1 + cot 2 θ ) = 1

Trigonometry

SOHCAHTOA -Introduction to Trigonometry
Trigonometric ratios and their Relation
Trigonometry for specific angles
Complementary angles in Trigonometry
Trigonometric Equations