Answer:
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TRIGONOMETRY IDENTITIES TO BE USED IN THE QUESTION :-
For any right angled triangle with one angle α ,
- [tex]\cos (90 - \alpha ) = \sin \alpha[/tex] or [tex]\sin(90 - \alpha ) = \cos\alpha[/tex]
- [tex]cosec \: (90 - \alpha ) = \sec\alpha[/tex] or [tex]\sec(90 - \alpha ) = cosec\:\alpha[/tex]
SOME GENERAL TRIGNOMETRIC FORMULAS :-
- [tex]\sin \alpha = \frac{1}{cosec \: \alpha }[/tex] or [tex]cosec \: \alpha = \frac{1}{\sin \alpha }[/tex]
- [tex]\cos \alpha = \frac{1}{\sec \alpha }[/tex] or [tex]\sec \alpha = \frac{1}{\cos \alpha }[/tex]
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Now , lets come to the question.
In a right angled triangle , let one angle be α (in place of theta) .
So , lets solve L.H.S.
[tex]\cos (90 - \alpha ) \times cosec(90 - \alpha )[/tex]
[tex]=> sin\alpha \times \sec\alpha[/tex]
[tex]=> \sin\alpha \times \frac{1}{\cos\alpha }[/tex]
[tex]=> \frac{\sin\alpha }{\cos\alpha }[/tex]
[tex]=> \tan\alpha[/tex] = R.H.S.
∴ L.H.S. = R.H.S. (Proved)
