# Trigonometry

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

Note by Alok Patel
4 years, 6 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.

- 4 years, 6 months ago

Thank you to help me in problem solving

- 4 years, 6 months ago

You are welcome!

- 4 years, 6 months ago

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

- 4 years, 6 months ago