Question

prove that cotA + Tan A=cosecA. secA​

Answer 1

Answer:

Step-by-step explanation:

We know that Cot A=Cos A/Sin A

                       Tan A=Sin A/Cos A

Cos A/Sin A+Sin A/Cos A

= Cos^2A+Sin^2A/Sin A Cos A

Cos^2A+Sin^2A=1

=1/Sin A Cos A

=1/Sin A* 1/Cos A

=Cosec A * Sec A

Hence Proved

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Answer 2

To prove:-

$\rm \cot A + \tan A = cosec \ A \sec A$

Solution:-

$\rm \cot A + \tan A = cosec \ A \sec A$

We know that,

$\cot A = \dfrac{\cos A}{\sin A} \quad and \quad \tan A = \dfrac{\sin A}{\cos A}$

Therefore,

$\rm \dashrightarrow\dfrac{\cos A}{\sin A}+ \dfrac{\sin A}{\cos A}= cosec \ A \sec A$

$\rm \dashrightarrow\dfrac{(\cos A)(\cos A) + (\sin A)(\sin A)}{(\sin A)(\cos A)}= cosec \ A \sec A$

$\rm \dashrightarrow\dfrac{\cos^2 A+ \sin^2 A}{(\sin A)(\cos A)}= cosec \ A \sec A$

We know that,

$\sin^2 A + \cos^2 A = 1$

Therefore,

$\rm \dashrightarrow\dfrac{1}{(\sin A)(\cos A)}= cosec \ A \sec A$

$\rm \dashrightarrow\dfrac{1}{\sin A} \times \dfrac{1}{\cos A}= cosec \ A \sec A$

Since, $\sin A = \dfrac{1}{\rm cosec \ A} \ and \ \cos A = \dfrac{1}{\sec A}$

Therefore,

$\rm \dashrightarrow cosec \ A \sec A= cosec \ A \sec A$

Hence, Proved.