Question

The value of \frac{\cot \theta}{\cot \theta-\cot 3 \theta}+\frac{\tan \theta}{\tan \theta-\tan 3 \theta}
cotθ−cot3θ
cotθ
​
+
tanθ−tan3θ
tanθ
​


Solution:

To simplify, \frac{\cot \theta}{\cot \theta-\cot 3 \theta}+\frac{\tan \theta}{\tan \theta-\tan 3 \theta}
cotθ−cot3θ
cotθ
​
+
tanθ−tan3θ
tanθ
​


Take LCM, the equation becomes,

=\frac{[\cot \theta \times(\tan \theta-\tan 3 \theta)][\tan \theta \times(\cot \theta-\cot 3 \theta]}{(\cot \theta-\cot 3 \theta) \times(\tan \theta-\tan 3 \theta)}=
(cotθ−cot3θ)×(tanθ−tan3θ)
[cotθ×(tanθ−tan3θ)][tanθ×(cotθ−cot3θ]
​


=\frac{[\cot \theta \times(\tan \theta-\tan 3 \theta)][\tan \theta \times(\cot \theta-\cot 3 \theta]}{(\cot \theta \times \tan \theta)-(\cot 3 \theta \times \tan \theta)+(\cot \theta \times \tan \theta)-(\tan 3 \theta \times \cot \theta)}=
(cotθ×tanθ)−(cot3θ×tanθ)+(cotθ×tanθ)−(tan3θ×cotθ)
[cotθ×(tanθ−tan3θ)][tanθ×(cotθ−cot3θ]
​


Since \cot \theta \times \tan \theta=1cotθ×tanθ=1

=\frac{1-(\cot \theta \times \tan 3 \theta)+1-(\tan \theta \times \cot 3 \theta)}{(1)-(\cot 3 \theta \times \tan \theta)+(1)-(\tan 3 \theta \times \cot \theta)}=
(1)−(cot3θ×tanθ)+(1)−(tan3θ×cotθ)
1−(cotθ×tan3θ)+1−(tanθ×cot3θ)
​


=\frac{2-(\cot \theta \times \tan 3 \theta)-(\tan \theta \times \cot 3 \theta)}{(2)-(\cot 3 \theta \times \tan \theta)-(\tan 3 \theta \times \cot \theta)}=
(2)−(cot3θ×tanθ)−(tan3θ×cotθ)
2−(cotθ×tan3θ)−(tanθ×cot3θ)
​


Since numerator and denominator are equal, \frac{\cot \theta}{\cot \theta-\cot 3 \theta}+\frac{\tan \theta}{\tan \theta-\tan 3 \theta}=1
cotθ−cot3θ
cotθ
​
+
tanθ−tan3θ
tanθ
​
=1​

Answer 1

Answer:

Which class you please tell me

Step-by-step explanation:

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