cot^-1(√1+sinx+√1-sinx/√1+sinx-√1-sinx) || cot inverse{√ 1+ sinx+ √ 1- sinx}/{ √1+sinx-√1-sinx}

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concepts & Methods

concepts & Methods

3 жыл бұрын

cot^-1(√1+sinx+√1-sinx/√1+sinx-√1-sinx)
cot inverse{√ 1+ sinx+ √ 1- sinx}/{ √1+sinx-√1-sinx}
Maths, Inverse Trigo , cot inverse{√ 1+ sinx+ √ 1- sinx}/{ √1+sinx-√1-sinx}
`cot^(-1)((sqrt(1+sinx)+sqrt(1-sinx))/(sqrt(1+sinx)-sqrt(1-sinx)))=x/2`
#cot^-1(√1+sinx+√1-sinx/√1+sinx-√1-sinx)
#cot inverse{√ 1+ sinx+ √ 1- sinx}/{ √1+sinx-√1-sinx}
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