Question

120 question . please answer thid

Answer 1

HELLO DEAR,



(tanA + tanB)/(cotA + cotB)


⇒ [ tanA + tanB) / (tanA + tanB) /tanA.tanB ]
∴ [ cotФ = 1/tanФ ]

⇒(tanA + tanB ) * (tanA . tanB ) / (tanA + tanB )

⇒tanA . tanB
∴ [ (tanA + tanB ) is cancelled ]


I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

hey here is your answer
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$l.h.s = \frac{\tan(a) + \tan(b) }{ \cot(a) + \cot(b) } \\ \\ = \frac{ \frac{ \sin(a) \cos(b) + \ \sin(b) \cos(a) }{ \cos( a) \cos(b) } }{ \frac{ \cos(a) \sin(b) + \cos(b) \sin(a) }{ \sin(a) \sin(b) } } \\ \\ = \frac{ \frac{1}{{ \cos( a) \cos(b) }} }{ \frac{1}{{ \sin(a) \sin(b) }} } \\ \\ = \frac{{ \sin(a) \sin(b) }}{{ \cos( a) \cos(b) }} \\ \\ = \tan(a) \tan(b) = r.h.s$
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Hope it may helpful to you
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