Question

1/tanA + cotA
= sinA. COSA​

Answer 1

Step-by-step explanation:

We have,,

$\sf{L.H.S.=\frac{1}{\tan(A)+\cot(A)}}$

$\sf{=\frac{1}{\frac{\sin(A)}{\cos(A)}+\frac{\cos(A)}{\sin(A)}}}$

$\sf{=\frac{1}{\frac{\sin^2(A)+\cos^{2}(A)}{\sin(A).\cos(A)}}}$

$\sf{=\frac{1}{\frac{1}{\sin(A).\cos(A)}}}$

$\sf{\sf{={\boxed{\sf\sin(A).\cos(A) }}= R.H.S.}}$

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Answer 2

Step-by-step explanation:

1/tanA+cotA

1/sin/cos+cos/sin

1/sin²+cos²/cos.sin

cos.sin/sin²+cos ²

(sin²+cos ²=1)

so,

proved

1/tanA+cotA= cos.sin