Trigonometry

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

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`2 \times 3`	$2 \times 3$
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Comments

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 - 7 years, 6 months ago

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 - 7 years, 6 months ago

Thank you to help me in problem solving

Alok patel - 7 years, 6 months ago

You are welcome!

Ajay Maity - 7 years, 6 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$