Q27 | Differentiate tan^(-1)((1+cosx)/sinx) | CBSE Previous Years Questions | Class 12
Q) If y = tan -¹(1-cosx /sinx) then dy/dx is #cbse2026 #maths #cbse #cbseboard #cbseclass12
`tan^(- 1)[(cosx)/(1+sinx)]` の `x` に関する微分を求めます。
Differentiate tan ^(-1) ((1+cosx)/sinx ) with respect to x | tan inverse 1 + cos x by sin x
Q) If 𝑦=tan^(−1) ((1−cos𝑥)/sin𝑥), then 𝑑𝑦/𝑑𝑥 #cbse2026 #maths #class12maths #cbse #cbseboard
Differentiate tan^-1((1+cosx)/sinx)) then dy/dx / tan inverse((1+cosx)/sinx)) / Differentiation
If `y = tan^(-1) ((1 - cos x)/(sin x)) " then "(dy)/(dx)=` ?
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Q13 | Differentiate tan^(-1)(cosx/(1+sinx ))
If y = tan^-1 { ( cos x - sin x )/( cos x + sin x ) , find dy/dx #maths #cbse2026 #cbse2026 #cbse
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If ` y = tan^(-1)((cos x)/(1 + sin x)), "then" (dy)/(dx) ` is equal to
