मराठी

Show that 2tan-1{tan α2⋅tan(π4-β2)}=tan-1 sinαcosβcosα+sinβ - Mathematics

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प्रश्न

Show that `2tan^-1 {tan  alpha/2 * tan(pi/4 - beta/2)} = tan^-1  (sin alpha cos beta)/(cosalpha + sinbeta)`

बेरीज

उत्तर

L.H.S. = `tan^-1  (2tan  alpha/2 * tan (pi/4 - beta/2))/(1 - tan^2  alpha/2 tan^2 (pi/4 - beta/2))`  ......`("since"  2 tan^-1x = tan^-1  (2x)/(1 - x^2))`

= `tan^-1  (2tan  alpha/2  (1 - tan  beta/2)/(1 + tan  beta/2))/(1 - tan^2  alpha/2  ((1 - tan  beta/2)/(1 + tan  beta/2))^2)`

= `tan^-1  (2tan  alpha/2 (1 - tan^2  beta/2))/((1 + tan  beta/2)^2 - tan^2  alpha/2 (1 - tan  beta/2)^2)`

= `tan^-1  (2tan  alpha/2 (1 - tan^2  beta/2))/((1 + tan^2  beta/2)(1 - tan^2  alpha/2) + 2   beta/2 (1 + tan^2  alpha/2))`

= `tan^-1  ((2tan  alpha/2)/(1 + tan^2  alpha/2) - (1 - tan^2  beta/2)/(1 + tan^2  beta/2))/((1 - tan^2  alpha/2)/(1 + tan^2  alpha/2) + (2tan  beta/2)/(1 + tan^ beta/2))`

= `tan^-1  ((sin alpha cos beta)/(cos alpha + sin beta))`

= R.H.S.

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पाठ 2: Inverse Trigonometric Functions - Solved Examples [पृष्ठ २७]

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एनसीईआरटी एक्झांप्लर Mathematics [English] Class 12
पाठ 2 Inverse Trigonometric Functions
Solved Examples | Q 20 | पृष्ठ २७

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