plz gys solve this question
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Given sin theta + cos theta = root 2.
Squaring on both sides, we get
(sin theta + cos theta)^2 = (root 2)^2
We know that (a+b)^2 = a^2+b^2+2ab.
(sin^2 theta + cos^2 theta + 2 sin theta cos theta) = 2
We know that sin^2 theta + cos^2 theta = 1
1 + 2 sin theta cos theta = 2
2 sin theta cos theta = 1
sin theta cos theta = 1/2 ---- (1)
Given tan theta + cot theta
= sin theta/cos theta + cos theta/sin theta
= sin^2 theta + cos^2 theta/sin theta cos theta
= 1/1/2 (From (1) - sin theta cos theta = 1/2)
= 2.
Hope this helps!
Squaring on both sides, we get
(sin theta + cos theta)^2 = (root 2)^2
We know that (a+b)^2 = a^2+b^2+2ab.
(sin^2 theta + cos^2 theta + 2 sin theta cos theta) = 2
We know that sin^2 theta + cos^2 theta = 1
1 + 2 sin theta cos theta = 2
2 sin theta cos theta = 1
sin theta cos theta = 1/2 ---- (1)
Given tan theta + cot theta
= sin theta/cos theta + cos theta/sin theta
= sin^2 theta + cos^2 theta/sin theta cos theta
= 1/1/2 (From (1) - sin theta cos theta = 1/2)
= 2.
Hope this helps!
harshsonker12p0a8z1:
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Answered by
1
given sin∅°+cos∅° = √2
divide each term by sin ∅
1+cot∅ = √2/sin∅
cot ∅ = √2-sin∅/sin∅
@@@@@@@@@@@@@@@@@@
tan∅+cot∅
= 1+cot²∅/cot∅
substitute cot∅ in the eqtn
divide each term by sin ∅
1+cot∅ = √2/sin∅
cot ∅ = √2-sin∅/sin∅
@@@@@@@@@@@@@@@@@@
tan∅+cot∅
= 1+cot²∅/cot∅
substitute cot∅ in the eqtn
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