In this section we will discuss Trigonometric ratios and their relation
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,
1) Csc θ = 1 / sin θ or sin θ = 1 / cscθ
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 Some solved examples on trigonometric ratios and their relation :
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. Solution : Using definition of trigonometric ratios, we have
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,
AC2 = AB 2 + BC2
(2k) 2 = AB2 + (1k) 2
4k2 = AB2 + k 2
AB2 = 4k 2 - k 2 ⇒ AB= √(3) k
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? Solution : We know ∠ A = ∠ C
AB = BC [ Sides opposite to equal angles are equal. ]
Cos A = AB / AC
= BC / AC [ substitution ]
= Cos C
Cos A = Cos C. Trigonometry
