यदि ` cos A = (24)/25 , ` तो निम्नलिखित के मान ज्ञात कीजिए जबकि ` 0 lt A lt pi/2`
(i) `sin 2A` (ii) ` cos 2A`
(iii) `tan 2A`

Explanation

` cos A = 24/25 ` जबकि A प्रथम चतुर्थाश में है ।
` :. Sin^(2) A = 1 - cos^(2) A `
` = 1 - (24/25)^(2) = 49/625`
`rArr sin A = 7/25` ( A प्रथम चतुर्थाश में है )
` tan A = (sinA)/(cosA) = (7/25)/(24/25)= 7/24 `
अब (i) ` sin 2 A sin A cosA `
` = 2* 7/(25)* 24/25= 336/625`
(ii) ` cos 2A = 2 cos^(2) A-1`
` = 2 * (24/25)^(2) -1`
` = (1152 - 625 )/625 = 527/625`
(iii) ` tan 2A = (2 tanA)/(1-tan^(2) A)`
` = (2(7/24))/(1-(7/24)^(2))`
` = (2 xx 7 xx 24)/(576 - 49) = 336/527`