Question:
prove that sin ^ 1 5/13 cos^1 3/5 = tan ^1 (63/16)
Answer:
Given, sin-1 (5/13) + cos-1 (3/5)
= tan-1 [(5/13)/√{1 - (5/13)2 }] + tan-1 [√{1 - (3/5)2 }/(3/5)] [ since sin-1 x = tan-1 {x/√(1 - x)2 } and cos-1 x = tan-1 {√(1 - x)2 /x} ]
= tan-1 [(5/13)/√{1 - 25/169}] + tan-1 [√{1 - 9/25}/(3/5)]
= tan-1 [(5/13)/√{(169 - 25)/169}] + tan-1 [√{(25 - 9)/25}/(3/5)]
= tan-1 [(5/13)/√{144/169}] + tan-1 [√{16/25}/(3/5)]
= tan-1 [(5/13)/(12/13)] + tan-1 [(4/5)/(3/5)]
= tan-1 (5/12) + tan-1 (4/3)
= tan-1 {(5/12 + 4/3)/(1 - 5/12 * 4/3)
= tan-1 [{(5*1 + 4*4)/12}/(1 - 20/36)]
= tan-1 [{(5 + 16)/12}/(1 - 20/36)]
= tan-1 [(21/12)/{(36 - 20)/36}]
= tan-1 [(21/12)/(16/36)]
= tan-1 {(21 * 36)/(12 * 16)}
= tan-1 {(21 * 6)/(2 * 16)}
= tan-1 {(21 * 3)/16}
= tan-1 (63/16)
So, sin-1 (5/13) + cos-1 (3/5) = tan-1 (63/16)
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