Question

Prove that tanA+cotA = 2

Answer 1

bro I am saying that it is not be equal to 2 I think it must be equal to

Answer 2

Solution :

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Given & To Prove :

⇒ Tan A + Cot A = 2

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⇒ Tan A + $\frac{1}{Tan A}$ = 2

⇒ $\frac{Tan^2A + 1}{tan A} = 2$ = 2

⇒ Tan²A + 1 = 2Tan A

⇒ Tan²A - 2Tan A + 1 = 0

⇒ (Tan A - 1)² = 0

⇒ Tan A - 1 = 0

⇒Tan A = 1

We know that,

Tan 45° = 1

Hence,  A = 45°

So,

Under the condition ,

If A = 45°

Then, Tan  A + cot A = 2

Aliter,.

(We know that,

Tan A = $\frac{sin A}{cos A}$

&

Cot A = $\frac{cos A}{sin A}$ )

So,

⇒ $\frac{sin A}{cos A} + \frac{cos A}{sin A}$ = 2

⇒ $\frac{sin^2 A + cos ^2A}{sinAcosA}$ = 2

⇒ $\frac{1}{sin A cos A}$ = 2

⇒ $1 = 2sinAcosA$

Again if A = 45°

⇒ 1 = $2( \frac{1}{ \sqrt{2} }) ( \frac{1}{ \sqrt{2} } )$

⇒ 1 = $2 ( \frac{1}{2} )$

⇒ 1 = 1

LHS = RHS

But, if A ≠ 45°

It wouldn't be possible !
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                                 Hope it Helps !!

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