If A + B + C = frac{pi }{2} ,then value of ta | vedclass

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Question

We have,

$A + B + C =90$ (degree)

Now,

$A+B=90-C \ldots \ldots .(1)$

Then,

$	an (A+B)=	an (90-C)(	ext { from }(1))$

or, $	an (A+

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$1$

Vedclass · Answer

We have,

$A + B + C =90$ (degree)

Now,

$A+B=90-C \ldots \ldots .(1)$

Then,

$	an (A+B)=	an (90-C)(	ext { from }(1))$

or, $	an (A+B)=	an (90-C) \ldots \ldots(2)$

$As$

$	an (x+y)=(	an x+	an y) /(1-	an x 	an y) \ldots \ldots(3)$

and, $	an (90-x)=\cot x \ldots \ldots(4)$

$\cup \operatorname{sing}(3)$ and (4) in $(2),$ we get

$(	an A+	an B) /(1-	an A 	an B)=\cot C$

or, $(	an A+	an B) /(1-	an A 	an B)=1 / 	an C$

or, $(	an C)(	an A+	an B)=(1-	an A 	an B)$

or, $(	an C 	an A)+(	an B 	an C)=(1-	an A 	an B)$

or, tan $A$ tan $B +	an B 	an C +	an C 	an A =1$

Therefore, we have

$	an A 	an B +	an B 	an C +	an C 	an A =1$

If $A + B + C = \frac{\pi }{2}$ ,then value of $tanA\,\, tanB + tanB\,\, tanC + tanC\,\, tanA$ is

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