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

Note by Alok Patel 3 years, 10 months 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.

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Thank you to help me in problem solving

You are welcome!

In the right hand side of the equation, do you mean \(\text{sec A csc A + 1}\) or \(\text{sec A (csc A +1)}\)?

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## Comments

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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.

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Thank you to help me in problem solving

Log in to reply

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 \(\text{sec A (csc A +1)}\)?

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