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The Trigonometric-ratios sin θ , cos θ, and tan θ of an angle θ are very closely connected by relation. If any one of them is known the other two can be easily calculated.

Sin θ = BC / AB Cos θ = AC / AB and tan θ = BC / AC

tan θ = BC / AC

tan θ = ( BC ÷ AB ) / (AC ÷ AB) [ Divide both numerator and denominator by AB]

tan θ = Sin θ / Cos θ [ since BC / AB = Sin θ and AC / AB = Cos θ ]

It is clear from the definitions of the Trigonometric-ratios that for any acute angle θ , we have,

2) Sec θ = 1 / cos θ or cos θ = 1 / sec θ

3) Cot θ = 1 / tan θ or tanθ = 1 / cotθ

4) Cot θ = cos θ / sin θ

5) tan θ cot θ = 1

1) In Δ ABC, right angled at A, if AB = 12, AC = 5 and BC = 13 , find all the six Trigonometric-ratios of angle B.

Sin B = AC / BC = 5 / 13

Cos B = AB / BC = 12 / 13

Tan B = AC / AB = 5 / 12

Csc B = BC / AC = 13 / 5

Sec B = BC / AB = 13 / 12

Cot B = AB / AC = 12 / 5

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2) If cosecA = 2, find the value of tanA.

Since CosecA=AC/BC

2 / 1 = AC / BC

So, Let AC = 2k, BC = 1k

By Pythagorian Theorem,

AC

(2k)

AB

So, tanA = BC / AB

= 1k /√(3) k

tanA = 1/√(3)

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3) Δ ABC is right angled at B and ∠A = ∠C. Is cosA = cosC?

AB = BC [ Sides opposite to equal angles are equal. ]

Cos A = AB / AC

= BC / AC [ substitution ]

= Cos C

Cos A = Cos C.

• SOHCAHTOA -Introduction to Trigonometry

• Trigonometric ratios and their Relation

• Trigonometry for specific angles

• Complementary angles in Trigonometry

• Trigonometric Equations

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