Question

tanΘ-tanΦ=x and cotΦ-cotΘ=y
Then Prove That cot(Θ-Φ)=$\frac{1}{x} +\frac{1}{y}$

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Answer 1

Step-by-step explanation:

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L. H. S multiplied by 0=0

R. H. S multiplied by 0=0

Hence proved

L. H. S=R. H. S

(proved)

Answer 2

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tanθ-tanφ = x

=> $\frac{1}{cotθ}$-$\frac{1}{cotφ}$

=> $\frac{cotφ-cotθ}{cotθ.cotφ}$ = $x$

=> $\frac{y}{cotθ.cotφ}$ = $x$

=> ${cotθ.cotφ}$ = $\frac{y}{x}$

And,

cotφ-cotθ = y

$\red{\bold{\underline{\underline{LHS:}}}}$

cot(θ-φ)

= $\frac{cotθ.cotφ + 1 }{cotφ-cotθ}$

= $\frac{y/x + 1}{y}$

= $\frac{(y+x)/x}{y}$

= $\frac{y+x}{xy}$

= $\frac{y}{xy}$ + $\frac{x}{xy}$

$❥Hence\: Proved$

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