यदि ` 3 cot A = 4 ` तो जांच कीजिए कि
`(1 - tan ^(2) A)/(1 + tan^(2) A) = cos ^(2) A - sin^(2)A ` है या नहीं |

Explanation

` cot A = (4) `
`rArr cot A = (4)/(3)`
अब ` cot A = ("आधार ")/("लम्ब") = (4)/(3)`
` (##NTN_HIN_MAT_X_C08_E06_008_S01.png" width="80%">
` Delta ABC "में" , angle B = 90^(@)`
आधार AB = 4K , लम्ब BC = 3k
` therefore AC^(2) = AB^(2) + BC^(2) = (4k)^(2) + (3k)^(2)`
` = 16 k^(2) + 9k^(2) = 25k^(2)`
`rArr Ac = 5k `
अब ` tan A = ("आधार ")/("लम्ब") = (BC)/(AB)( = (3k)/(4k) = (3)/(4)`
` cos A = ("आधार ")/("कर्ण") (AB)/(AC) = (4k)/(5k) = (4)/(5)`
` sin A = ("लम्ब")/("कर्ण") = (BC)/(AC) = (3k)/(5k) = (3)/(5)`
अब ` (1 - tan^(2) A )/(1 + tan^(2) A ) = (1 - ((3)/(4))^(2))/(1 + ((3)/(4))^(2)) `
तथा ` cos^(2) A - sin^(2) A = ((4)/(5))^(2) - ((3)/(5))^(2)`
` = (16)/(25)-(9)/(25) = (16 -9)/(25) = (7)/(25)`
` therefore (1 - tan^(2) A)/(1 + tan ^(2)A) = cos ^(2)A - sin^(2) A `