यदि ` cot theta = (7)/(8) ` ,तो
(i) ` ((1 + sin theta )(1 - sin theta))/((1 + cos theta )( 1 - cos theta))`,
(ii) ` cot^(2)` का मान निकालिए |

Explanation

`cot theta = (7)/(8)`
` cot theta = ("आधार ")/("लम्ब") = (7)/(8)`

माना `Delta ABC " में" angle B = 90^(@) " तथा" angle A = theta `
माना आधार AB = 7k
तथा लम्ब BC = 8k
अब `AC^(2) = AB^(2) + BC^(2) = (7k)^(2) + (8k)^(2)`
= ` 49k^(2) + 64K^(2) = 113k^(2)`
` rArr AC sqrt(113k)`
अब ` sin theta = ("लम्ब")/("कर्ण") = (BC)/(AC) = (8k)/(sqrt(113k)) = (8)/(sqrt(113))`
तथा ` cos theta = ("आधार ")/("कर्ण") = (AB)/(AC)= (7k)/(sqrt(113k)) = (7)/(sqrt(113))`
(i) `((1 + sin theta )(1 - sin theta))/((1 + cos theta )(1 - cos theta)) = ((1 +(8)/(sqrt(113)))(1 - (8)/(sqrt(113))))/((1 +(7)/(sqrt(113)))(1 - (7)/(sqrt(113))))`
`((1)^(2) - ((8)/(sqrt(113)))^(2) )/((1)^(2) - ((7)/(sqrt(113)))^(2) )=(1 - (64)/(113))/(1-(49)/(113))`
`(113 - 64)/(113 - 49) = (49)/(64)`
(ii) ` cot^(2) theta ((7)/(8))^(2) = (49)/(64)`