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(Barré) #1

9 ⋅5 Trigonometric identity


(i)(sinθ)^2 (cosθ)^2 =

2 2
̧
¹

·
̈
©

§
̧ 
¹

·
̈
©

§
OP

OM
OP

PM

= 2


2
2

2
OP

OM
OP

PM
 = 2

2 2
OP

PM OM
= 2

2
OP

OP
[ by the formula of Pythagoras ]

1
or,(sinθ)^2 (cosθ)^2 1


? sin^2 θcos^2 θ 1

Remark : For integer indices n we can write sinnθ for (sinθ)n and cosnθ for (cosθ)n.


(ii) sec^2 θ=(secθ)^2 =

2
̧
¹

·
̈
©

§
OM

OP

= 2

2

OM

OP
= 2

2 2

OM

PM OM
[OP is the hypotenuse of right angled 'POM]

= 2

2
2

2
OM

OM
OM

PM


=


2

(^1) ̧
¹
·
̈
©

OM
PM
= 1 (tanθ)^2 = 1 tan^2 θ
? sec^2 θ 1 tan^2 θ
or, sec^2 θtan^2 θ 1
or, tan^2 θ sec^2 θ 1
(iii)cosec^2 θ (cosecθ)^2 =
2
̧
¹
·
̈
©
§
PM
OP
= 2
2
PM
OP
= 2
2 2
PM
PM OM
[i is the hypotenuse of right-angled 'POM]
= 2
2
2
2
PM
OM
PM
PM
 =
2
(^1) ̧
¹
·
̈
©
§

PM
OM
= 1 (cotθ)^2 = 1 cot^2 θ
? cosec^2 θcot^2 θ 1 and cot^2 θ cosec^2 θ 1


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