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प्रश्न
Prove that `2 tan^(-1) (1/3) + cos^(-1) (3/5) = pi/2`.
उत्तर
Using the double-angle formula for inverse tangent:
L.H.S. = `2 tan^(-1) (1/3) + cos^(-1) (3/5)`
= `tan^(-1)[(2 xx 1/3)/(1 - 1/3^2)] + cos^(-1) (3/5) ...[2tan^(-1)x = tan^(-1)((2x)/(1 - x^2))]`
= `tan^(-1)[(2 xx 1/3)/(1 - 1/3^2)] + tan^(-1)sqrt((1 - 9/25)/(3/5)) ...[cos^(-1)x = tan^(-1)(sqrt(1 - x^2)/x)]`
= `tan^(-1)[(2/3)/(1 - 1/9)] + tan^(-1) sqrt(((25 - 9)/25)/(3/5))`
= `tan^(-1)[(2/3)/((9 - 1)/9)] + tan^(-1)sqrt((16/25)/(3/5))`
= `tan^(-1)((2/3)/(8/9)) + tan^(-1)((4/5)/(3/5))`
= `tan^(-1)(2/3) xx(9/8) + tan^(-1)(4/5) xx (5/3)`
= `tan^(-1)(3/4) + tan^(-1)(4/3) ...[tan^(-1)a + tan^(-1)b = tan^(-1)((a + b)/(1 - ab))", if " ab < 1]`
= `tan^(-1)((3/4 + 4/3)/(1 - (3/4 xx 4/3)))`
= `tan^(-1)(((9 + 16)/12)/(1 - 1))`
= `tan^(-1)(((25)/12)/(0))`
= `tan^(-1)(0) ...(tan0° = pi/2)`
= `pi/2`
= R.H.S.
L.H.S. = R.H.S.
