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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) sin^{2}θ + cos^{2}θ= 12) sin ^{2}θ = 1 - cos^{2}θ3) cos ^{2}θ = 1 - sin ^{2}θ 4) 1 + tan ^{2}θ = sec^{2}θ5) tan ^{2}θ = sec^{2}θ - 16) sec ^{2}θ - tan^{2}θ = 1 7) 1 + cot ^{2}θ = csc^{2}θ 8) cot ^{2}θ = csc^{2}θ - 19) csc ^{2}θ - cot^{2}θ = 110) 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.

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

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 / cos

= 2 sec

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

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2) Prove that : cos θ / ( 1- sin θ ) = ( 1 + sin θ) / cos θ

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

[ using this identity (a + b)(a – b ) = a

= cos θ ( 1 + sin θ) / cos

= (1 + sin θ) / cos θ

∴ cos θ / ( 1- sin θ ) = ( 1 + sin θ) / cos θ

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3) Cos

Consider Cos

= Cos

= cos

= 1 [ since sin

∴ Cos

• SOHCAHTOA -Introduction to Trigonometry

• Trigonometric ratios and their Relation

• Trigonometry for specific angles

• Complementary angles in Trigonometry

• Trigonometric Equations

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