Question

सिद्ध करा: `2 tan^(-1) (1/3) + cos^(-1) (3/5) = pi/2`.

Answer 1

व्यस्त स्पर्शिकेसाठी दुहेरी-कोन सूत्राचा वापर करून:

डावी बाजू = `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))", जर " 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`

= उजवी बाजू

डावी बाजू = उजवी बाजू