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

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Trigonometry

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

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