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Prove cosA-sinA+1/cosA+sinA-1=cosecA+cotA |
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Answer» LHS=cosA+sinA−1cosA−sinA+1\u200bdividing Nr and Dr by\xa0sinA\xa0we get,=sinAcosA\u200b+sinAsinA\u200b−sinA1\u200bsinAcosA\u200b−sinAsinA\u200b+sinA1\u200b\u200b=cotA+1−cosecAcotA−1+cosecA\u200b=cotA+1−cosecAcotA+cosecA−(cosec2A−cot2A)\u200b=cotA+1−cosecA(cotA+cosecA)(1−cosecA+cotA)\u200b=cotA+cosecA=RHS Multiplying the numerator and denominator with sinA :SinA [CosA – SinA +1]/SinA [CosA + SinA –1]SinACosA–Sin^2A+SinA/SinA [CosA +SinA –1]SinACosA+SinA –[1–Cos^2A]/SinA[CosA+SinA–1]SinACosA+SinA–[(1–CosA)(1+CosA)] /SinA[CosA+SinA–1]SinA[CosA+1]–[(1–CosA)(1+CosA)] /SinA [CosA + SinA–1](CosA+1)(SinA–1+CosA)/SinA (CosA + SinA–1)(CosA+1)(SinA+CosA—1)/SinA (CosA+SinA–1)(CosA+SinA–1) will get cancelledSo , CosA+1/SinA is leftDividing (CosA) and (1) individually by (SinA), we get CotA + CosecA Which is equal to R.H.SHence Proved |
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