1- A-tan Asin A-cos A-sin A tan A- A cos A

1+cotA+tanAsinA cosA=sinA+cotAsinA+tanAsinA cosA cotAcosA tanAcosA=sinA+cosAsinA #215;sinA+tanAsinA cosA cotAcosA sinAcosA #215;cosA =sinA+cosA+tanAsinA co ...

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Standard XII

Mathematics

Implicit Differentiation

1+ A+tan Asin...

Question

(1 + cot A + tan A) (sin A – cos A) = sin A tan A – cot A cos A

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Solution

$(1 + \cot A + \tan A) (\sin A - \cos A) = \sin A + \cot A \sin A + \tan A \sin A - \cos A - \cot A \cos A - \tan A \cos A = \sin A + \frac{\cos A}{\sin A} \times \sin A + \tan A \sin A - \cos A - \cot A \cos A - \frac{\sin A}{\cos A} \times \cos A$
$= \sin A + \cos A + \tan A \sin A - \cos A - \cot A \cos A - \sin A = \sin A \tan A - \cot A \cos A$

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Similar questions

Q. Prove that

(1 + cot A + tan A) (sin A - cos A) = sin A tan A - cot A cos A

Q. Prove the following trigonometric identities.

(1 + \cot A + \tan A) (\sin A - \cos A) = \frac{\sec A}{{cosec}^{2} A} - \frac{cosec A}{\sec^{2} A} = \sin A \tan A - \cot A \cos A

Prove the following identities:

$(1 + c o t A + t a n A) (s i n A - c o s A) = \frac{s e c A}{c o s e c^{2} A} - \frac{c o s e c A}{s e c^{2} A} = s i n A t a n A - c o t A c o s A$

Show that:

(1 + cot A + tan A) (sin A - cos A) = \frac{sec A}{{cosec}^{2} A} - \frac{cosec A}{{sec}^{2} A} = sin A tan A - cot A cos A

Q. Prove:

(1 + c o t A + t a n A) (s i n A - c o s A) = s i n A t a n A - c o t A c o s A

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(1 + cot A + tan A) (sin A – cos A) = sin A tan A – cot A cos A

1+cotA+tanAsinA-cosA=sinA+cotAsinA+tanAsinA-cosA-cotAcosA-tanAcosA=sinA+cosAsinA×sinA+tanAsinA-cosA-cotAcosA-sinAcosA×cosA =sinA+cosA+tanAsinA-cosA-cotAcosA-sinA=sinAtanA-cotAcosA