# 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

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tan A + cot A = 5

Therefore

There for ( tan

Therefore ( tan

Therefore tan

Threfore tan

Therefore tan

**Squaring both sides**Therefore

**(tan A + cot A)**^{2}=5^{2}There for ( tan

^{2}A + cot^{2}A + 2 x tan^{2}A x cot^{2}A) = 25Therefore ( tan

^{2 }A + cot^{2}A + 2 x tan^{2}Ax 1/tan^{2}A) = 25Therefore tan

^{2}A + cot^{2}A + 2 =25Threfore tan

^{2}A + cot^{2}A = 25-2Therefore tan

^{2}A + cot^{2}A = 23- 1

(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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