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Prove the following statements. tanA/1-cotA+cotA/1-tanA=(secA.cosecA+1)

Note by Alok Patel 3 years, 1 month ago

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Here's how I did it:

\(\frac{tanA}{1-cotA}\) + \(\frac{cotA}{1-tanA}\)

= \(\frac{\frac{sinA}{cosA}}{1-\frac{cosA}{sinA}}\) + \(\frac{\frac{cosA}{sinA}}{1-\frac{sinA}{cosA}}\)

= \(\frac{sin^{2}A}{cosA(sinA-cosA)}\) + \(\frac{cos^{2}A}{sinA(cosA-sinA)}\)

= \(\frac{sin^{3}A-cos^{3}A}{sinAcosA(sinA-cosA)}\)

= \(\frac{(sinA-cosA)(sin^{2}A+sinAcosA+cos^{2}A)}{sinAcosA(sinA-cosA)}\)

= \(\frac{sin^{2}A+sinAcosA+cos^{2}A}{sinAcosA}\)

= \(\frac{1+sinAcosA}{sinAcosA}\)

= \(1+\frac{1}{sinAcosA}\)

= 1 + secAcosecA

Hope you understood. – Ajay Maity · 3 years, 1 month ago

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@Ajay Maity – Thank you to help me in problem solving – Alok Patel · 3 years, 1 month ago

@Alok Patel – You are welcome! – Ajay Maity · 3 years, 1 month ago

In the right hand side of the equation, do you mean \(\text{sec A csc A + 1}\) or \(\text{sec A (csc A +1)}\)? – 敬全 钟 · 3 years, 1 month ago

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TopNewestHere's how I did it:

\(\frac{tanA}{1-cotA}\) + \(\frac{cotA}{1-tanA}\)

= \(\frac{\frac{sinA}{cosA}}{1-\frac{cosA}{sinA}}\) + \(\frac{\frac{cosA}{sinA}}{1-\frac{sinA}{cosA}}\)

= \(\frac{sin^{2}A}{cosA(sinA-cosA)}\) + \(\frac{cos^{2}A}{sinA(cosA-sinA)}\)

= \(\frac{sin^{3}A-cos^{3}A}{sinAcosA(sinA-cosA)}\)

= \(\frac{(sinA-cosA)(sin^{2}A+sinAcosA+cos^{2}A)}{sinAcosA(sinA-cosA)}\)

= \(\frac{sin^{2}A+sinAcosA+cos^{2}A}{sinAcosA}\)

= \(\frac{1+sinAcosA}{sinAcosA}\)

= \(1+\frac{1}{sinAcosA}\)

= 1 + secAcosecA

Hope you understood. – Ajay Maity · 3 years, 1 month ago

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– Alok Patel · 3 years, 1 month ago

Thank you to help me in problem solvingLog in to reply

– Ajay Maity · 3 years, 1 month ago

You are welcome!Log in to reply

In the right hand side of the equation, do you mean \(\text{sec A csc A + 1}\) or \(\text{sec A (csc A +1)}\)? – 敬全 钟 · 3 years, 1 month ago

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