Math, asked by sk5697613, 3 months ago

sinA -cosA+1/sinA+cosA-1=1/secA-tanA. prove this question​

Answers

Answered by Bhaiyaji12
0

L.H.S = tanA−secA+1

tanA+secA−1

= (tanA+secA)−(sec 2A−tan 2A) ÷ tanA−secA+1

= (tanA+secA)−(secA−tanA)(secA+tanA) ÷ tanA−secA+1

= (tanA+secA).[1−secA+tanA] ÷ tanA−secA+1

= tanA+secA

= sinA ÷ cosA + 1 ÷ cosA = 1 + sinA ÷ cosA

= R.H.S

Hence Proved

Answered by sandy1816
0

 \frac{sina - cosa + 1}{sina + cosa - 1}  \\  \\  =  \frac{ \frac{sina - cosa + 1}{cosa} }{ \frac{sina + cosa - 1}{cosa} }  \\  \\  =  \frac{tana + seca - 1}{tana - seca + 1}  \\  \\  =  \frac{tana + seca - 1}{( {sec}^{2} a -  {tan}^{2}a) - (seca - tana) }  \\  \\  =  \frac{tana + seca - 1}{(seca - tana)(seca + tana - 1)}  \\  \\  =  \frac{1}{seca - tana}

Hopes it helps

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