Math, asked by harshitjogi, 4 months ago

tanA/1-cotA+cotA/1-tanA=1+secAcosecA​

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Answered by Thakurjanvi003
1

Answer:

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Step-by-step explanation:

hey hope it helps you i hv solved it

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Answered by sandy1816
0

 \frac{tanA}{1 - tanA}  +  \frac{cotA}{1 - tanA}  \\  \\  =  \frac{ \frac{cosA}{sinA} }{ \frac{cosA - sinA}{cosA} }  +  \frac{ \frac{sinA}{cosA} }{ \frac{sinA - cosA}{sinA} }  \\  \\  =  \frac{ {cos}^{2}A }{sinA(cosA - sinA)}  +  \frac{ {sin}^{2}A }{cosA(sinA - cosA)}  \\  \\  =  \frac{ {cos}^{2} A}{sinA(cosA - sinA)}  -  \frac{ {sin}^{2}A }{cosA(cosA - sinA)}  \\  \\  =  \frac{ {cos}^{3}A-  {sin}^{3} A }{sinAcosA(cosA - sinA)}  \\  \\  =  \frac{(cosA - sinA)(1 + sinAcosA)}{sinAcosA(cosA - sinA)}  \\  \\  =  \frac{1 + sinAcosA}{sinAcosA}  \\  \\  = 1 + secAcosecA

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