solve the given differetiation
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y=tan'(1+sinx/cosx)
y= tan'(cos²x/2+sin²x/2+2sinx/2+sin/2/cos²x/2-sin²x/2)
y= tan'[(cosx/2+sinx/2)²/cos²x/2-sin²x/2)]
y=tan'[cosx/2+sinx/2/(cosx/2+sinx/2)(cosx/2-sinx/2)]
y=tan'[whole ÷cosx/2]
y=tan'(1+tanx/2/1-tanx/2)
y=tan'(tanπ/4+x/2)
y=π/4+x/2
dy/dx=1/2
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