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प्रश्न
Prove that:
`cos^(-1) 4/5 + cos^(-1) 12/13 = cos^(-1) 33/65`
उत्तर १
Let `cos^(-1) 4/5 = x`. Then, `cos x = 4/5 => sin x = sqrt (1 - (4/5)^2) = 3/5`
`:. tan x = 3/4 => x = tan^(-1) 3/4`
`:. cos^(-1) 4/5 = tan^(-1) 3/4` ...(1)
Now let `cos^(-1) 12/13 = y` Then `cos y= 12/13 => sin y = 5/13`
`:. tan y = 5/12 => y = tan^(-1) 5/12`
`:. cos^(-1) 12/13 = tan^(-1) 5/12 --- 2`
Let `cos^(-1) 33/65 = z`. Then `cos z = 33/65 => sin z = 56/65`
`:. tan z = 56/33 => z = tan^(-1) 56/33`
`:. cos^(-1) 33/65 = tan^(-1) 56/33` ....(3)
Now, we will prove that:
L.H.S = `cos^(-1) 4/5 + cos^(-1) 12/13`
`= tan^(-1) 3/4 + tan^(-1) 5/12` [Using 1 and 2]
= `tan^(-1) (3/4 + 5/12)/(1 - 3/4 . 5/12)` ` " " [tan^(-1) x + tan^(-1) y = tan^(-1) (x + y)/(1-xy)]`
`= tan^(-1) (36+20)/(48-15)`
`= tan^(-1) 56/33`
`= tan^(-1) 56/33` [by(3)]
= R.H.S
उत्तर २
`cos^-1 4/5 + cos^-1 12/13`
` = tan^-1 (sqrt(5^2 - 4^2))/4 + tan^-1 sqrt(13^2 - 12^2)/12`
= `tan^-1 3/4 + tan^-1 5/12`
= `tan^-1 ((5/12 + 3/4)/(1 - 5/12 xx 3/4))` ...`[tan^-1x + tan^-1y = tan^-1((x + y)/(1 - x xx y))]`
= `tan^-1 (56/33)`
= `cos^-1 33/sqrt(56^2 + 33^2)`
= `cos^-1 33/65`
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