11 question pls
11 question pls (O. sin A + cosec A = 2;
find the value of sin2 A + cosec2
"f tan A + cot A = 5;
find the value of tan2 A + coe A.
Given : 4 sin 0 = 3 cos 0; find the
(i) sin 0
(ii) cos 0
tan A + cot A = 5 then by squaring on both sides we get,
25
25
- -3
tan A + cot A = 5
Squaring both sides
Therefore (tan A + cot A)2=52
There for ( tan2 A + cot2 A + 2 x tan2 A x cot2 A) = 25
Therefore ( tan2 A + cot2 A + 2 x tan2 Ax 1/tan2 A) = 25
Therefore tan2 A + cot2 A + 2 =25
Threfore tan2 A + cot2 A = 25-2
Therefore tan2 A + cot 2 A = 23
Squaring both sides
Therefore (tan A + cot A)2=52
There for ( tan2 A + cot2 A + 2 x tan2 A x cot2 A) = 25
Therefore ( tan2 A + cot2 A + 2 x tan2 Ax 1/tan2 A) = 25
Therefore tan2 A + cot2 A + 2 =25
Threfore tan2 A + cot2 A = 25-2
Therefore tan2 A + cot 2 A = 23
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(tan A + cot A)2??= 52? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? [from, (x+y)2?= x2?+ y2?+ 2xy]
=? tan2A + cot2A + 2*tan A*cot A = 25
=? tan2A + cot2A = 25 - 2*tan A*? ? 1? ???????
? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?tan A? ? ? ? ? ? ? ? ?
? tan2A + cot2A = 25-2 =?23
=? tan2A + cot2A + 2*tan A*cot A = 25
=? tan2A + cot2A = 25 - 2*tan A*? ? 1? ???????
? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?tan A? ? ? ? ? ? ? ? ?
? tan2A + cot2A = 25-2 =?23
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