cotA +cosecA-1/ CotA- cosecA+1 is equal to
Answers
Answer:
LHS=(cos A / sin A + 1/sin A - sin A / sin A) / (cos A / sin A - 1 / sin A + sin A / sin A)
= (cos A + 1 - sin A) / (cos A - 1 + sin A) * (cos A + 1 - sin A) / (cos A + 1 - sin A)
= (cos A + 1 - sin A) (cos A + 1 - sin A) / ( ((cos A - 1 + sin A) (cos A + 1 - sin A))
= (cos A + 1 - sin A)^2 / (cos^2 A - (1 - sin A)^2)
= (cos^2 A - 2 cos A (1 - sin A) + (1 - sin A)^2) / (cos^2 A - 1 + 2 sin A - sin^2 A)
= (cos^2 A - 2 cos A + 2 sin A cos A + 1 - 2 sin A + sin^2 A) / (1 - sin^2 A - 1 + 2 sin A - sin^2 A)
= (2 - 2 cos A + 2 sin A cos A - 2 sin A) / (2 sin A - 2 sin^2 A)
= 2 (1 - cos A) (1 - sin A) / (2 sin A (1 - sin A) )
= 1 - cos A / sin A
= RHS
(It could be quicker to use double angle formulae on half angles: 1 -/+ cos A are obvious there.)
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To Prove: cotA+cosecA−1cotA−cosecA+1=1+cosAsinA
I don’t think the expression is right. Because
LHS =cotA+cosecA−1cotA−cosecA+1
=cotA−1cotA+1 (Canceling positive and negative cosecA )
=1+cotA−1cotA
RHS =1+cosAsinA
=1+cotA (Since cosAsinA =cotA )
So, We can see that, −1cotA is already more expression there at LHS.
Also, Lets put A=30o
=> LHS => 3–√+2−13√−2+1=1+23√
=> RHS => 1+3–√
And, Lets put A=45o
=> LHS => [math]1 + {\sqrt 2} - 1 - {\sqr[/math]
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cotA+cosecA−1(cosec2A−cot2A)cotA−cosecA+1
(cotA+cosecA)−(cosecA+cotA)(cosecA−cotA)cotA−cosecA+1
(cotA+cosecA)(1−cosecA+cotA)cotA−cosecA+1
cotA+cosecA
cosAsinA+1sinA
1+cosAsinA
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See this…….
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Answer:
The Answer Is
Step-by-step explanation:
cotA + cosecA-1...
where,
= cosecA - 1 = cot A
then,
cotA - cosecA +1...
where,
= cosecA +1 = (-cotA)
so therefore ,
= cotA + cotA / cotA - (-cotA)
= 2cotA / 2cotA
= 1