Question

1 +
Prove the identity : tan A + cot A = sec A, cosec A​

Answer 1

Step-by-step explanation:

Consider L.H.S

tanA + cot A

= sinA/cos A + cos A / sin A

= sin²A + cos²A / sin A × cos A (cross multiplying )

= 1 / sin A × cos A ( sin²A + cos²A = 1 )

= 1/sinA × 1/cos A

= cosec A × sec A ( cosecA = 1/sin A , sec A = 1/cosA )

L.H.S = R.H.S

Hence Proved

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Answer 2

Correct Question:

Prove the identity :

$\sf{tan \: A + cot \: A = sec \: A \: . \: cosec \: A}$

Answer :-

Given to prove the identity :

• tan A + cot A = sec A . cosec A

L.H.S = $\sf{tan \: A + cot \: A}$

$\sf{ = \dfrac{sin \: A}{cos \: A} + \dfrac{cos \: A}{sin \: A}} \: \: \: \: \: \: \: \: \: \: \: \: {}_{.. \: ...} \: \: \bigg\lgroup \because \sf{ \red{tan \: A = \dfrac{sin \: A}{cos \: A}} \: \: and \: \: \red{cot \: A = \dfrac{cos \: A}{sin \: A}} \bigg \rgroup}$

$\sf{= \dfrac{{sin}^{2} \: A + {cos}^{2} \: A}{cos \: A \: . \: sin \: A}}$

$\sf{= \dfrac{1}{cos \: A \: . \: sin \: A}} \: \: \: \: \: \: \: \: \: \: {}_{... \: ..} \: \: \lgroup \because \sf{ \red{{sin}^{2} \: A + {cos}^{2} \: A = 1} \rgroup}$

$\sf{= sec \: A \: . \: cosec \: A = \bf{R.H.S}} \: \: \: \: \: \: \: \: \: \: \: \: {}_{.. \: ...} \: \: \bigg\lgroup \because \sf{ \red{sec \: A = \dfrac{1}{cos \: A}} \: \: and \: \: \red{cosec \: A = \dfrac{1}{sin \: A}} \bigg \rgroup}$

Hence, L.H.S = R.H.S

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Things to remember :-

• $\sf{tan \: A = \dfrac{sin \: A}{cos \: A}}$

• $\sf{cot \: A = \dfrac{cos \: A}{sin \: A}}$

• $\sf{cosec \: A = \dfrac{1}{sin \: A}}$

• $\sf{sec \: A = \dfrac{1}{cos \: A}}$

• $\sf{{sin}^{2} \: A + {cos}^{2} \: A = 1}$