Prove that:-1+cotA÷cosA + 1+tanA ÷ sinA = 2 (sec A + cosec A)

1.	Prove that:-1+cotA÷cosA + 1+tanA ÷ sinA = 2 (sec A + cosec A)
Answer» $\bigstar$ ANSWER: $\frac{1+cotA}{cosA}+\frac{1+tanA}{sinA}=2(secA+cscA)$ $\bigstar$ Step-by-step EXPLANATION: $\bigstar$ L.H.S $\implies\frac{1+cotA}{cosA}+\frac{1+tanA}{sinA}\\\\\implies \frac{1+\frac{cosA}{sinA} }{cosA}+\frac{1+\frac{sinA}{cosA} }{sinA}\\\\\implies \frac{sinA+cosA}{sinA.cosA}+\frac{cosA+sinA}{cosA.sinA}\\\\ \implies \frac{sinA}{sinA.cosA}+\frac{cosA}{sinA.cosA}+\frac{cosA}{cosA.sinA}+\frac{sinA}{cosA.sinA}\\\\ \implies \frac{1}{cosA}+\frac{1}{sinA}+\frac{1}{sinA}+\frac{1}{cosA} \\\\\\implies secA+cscA+ cscA+secA\\\\\implies 2secA+2cscA\\\\\implies 2(secA+cscA)$ $\bigstar$ R.H.S

Answer»

$\bigstar$ ANSWER:

$\frac{1+cotA}{cosA}+\frac{1+tanA}{sinA}=2(secA+cscA)$

$\bigstar$ Step-by-step EXPLANATION:

$\bigstar$ L.H.S

$\implies\frac{1+cotA}{cosA}+\frac{1+tanA}{sinA}\\\\\implies \frac{1+\frac{cosA}{sinA} }{cosA}+\frac{1+\frac{sinA}{cosA} }{sinA}\\\\\implies \frac{sinA+cosA}{sinA.cosA}+\frac{cosA+sinA}{cosA.sinA}\\\\ \implies \frac{sinA}{sinA.cosA}+\frac{cosA}{sinA.cosA}+\frac{cosA}{cosA.sinA}+\frac{sinA}{cosA.sinA}\\\\ \implies \frac{1}{cosA}+\frac{1}{sinA}+\frac{1}{sinA}+\frac{1}{cosA} \\\\\\implies secA+cscA+ cscA+secA\\\\\implies 2secA+2cscA\\\\\implies 2(secA+cscA)$

$\bigstar$ R.H.S

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