prove that
tan A/(1-cotA)+cotA/(1-tanA)=1+secAcosecA
Answers
Explanation:
tanA/(1-cotA) +cotA/(1-tanA)
=tanA/(1–1/tanA) +cotA/(1-tanA)
=tan^2A/(tanA-1) +cotA/(1-tanA)
=-tan^2A/(1-tanA) +cotA/(1-tanA)
=(-tan^2A+cotA)/(1-tanA)
=(-tan^2A+1/tanA)/(1-tanA)
=(-tan^3A+1)/tanA(1-tanA)
=(1-tan^3A)/tanA(1-tanA)
=(1-tanA)(1+tanA+tan^2A)/tanA(1-tanA)
=(1+tanA+tan^2A)/tanA
=(1+tan^2A+tanA)/tanA
=(sec^2A+tanA)/tanA
=sec^2A/tanA +tanA/tanA
=1/cos^2AtanA+1
=cosA/cos^2AsinA+1
=1/cosAsinA+1
=secAcosecA+1
How do I prove that tanA/1-cotA + cotA /1- tanA = 1+ secA. cosecA?
How do you prove that, tanA/1-cotA + CotA/1-tanA is equal to 1+tanA+cotA?
How do I show that tanA/ (1-cotA) +cotA/ (1-tanA) =secAcosecA+1?
How do I prove that: (1+sec2A) (1+sec4A) (1+sec8A) = tan8A.cotA?
If (tanA/1-cotA) + (cotA/ 1-tanA) =secAcosecA+1?
Easiest Approach…
Hope this helps…
I hope this answer helps .
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How do I prove: [math]\frac{\tan A}{1-\cot A} +\frac{\cot A}{1-\tan A} = \sec A\csc A + 1[/math]?
How do you prove that sinA(1+tanA)+cosA(1+cotA)=secA+cosecA?
How do you prove that (tan A) / (1-cot A) +cot A/ (1-tan A) =1+sec A cosec A?
tanA/(1-cotA) + cotA/(1-tanA)
sin^2 A/cosA(sinA-cosA) - cos^2 A/sinA(sinA-cosA)
1/(sinA-cosA) {(sin^3A - cos^3A)/sinA cosA}
1/(sinA-cosA){(sinA - cosA)(sin^2A+sinAcosA+cos^2A)/sinAcosA}
(1+sinAcosA)/sinAcosA
secAcosecA + 1
There could be shorter methods to be identified if we spend time, but I hit on the following proof.
LHS = (sinA/cosA)/(1-cosA/sinA) + (cosA/sinA)/(1-sinA/cosA)
=(sinA*sinA/cosA)/(sinA-cosA) + (cosA*cosA/sinA)/(cosA-sinA)
=(sin^2A/cosA)/(sinA-cosA) - (cos^A/sinA)/(sinA-cosA)
=(1/(sinA-cosA))*(sin^3A/sinAcosA - cos^3A/sinAcosA)
=(1/(sinA-cosA))*(Sin^3A-cos^3A)/sinAcosA
=(1/sinAcosA)*(sin^3A-cos^3A)/(sinA-cosA)
=(1/sinAcosA)(sin^2A+cos^2A+sinAcosA) {because (x^3-y^3)/(x-y)=x^2+y^2+xy}
= (1/sinAcosA) * (1+sinAcosA)
= (1/sinAcosA) + 1
= (1/sinA*1/cosA)+1
=secAcosecA+1=RHS
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See,in trigonometric questions it is always best to solve it in cosx and sinx
So, your left hand side becomes,
(sin^2x)÷((sinx-cosx)cosx) - (cos^2x)((sinx-cosx)sinx)
Now once you reached here, take least common factor or LCM, hence equation simplifies as
(sin^3x - cos^3x)÷((sinx-cosx)sinx.cosx)
In numerator use, a^3 - b^3 =(a-b)(a^2+b^2 + ab)
And sin^2x + cos^2x = 1.
So you will get your equation as,
(1+ sinx.cosx)(sinx-cosx)÷(sinx-cosx)(sinx.cosx)
So removing (sinx-cosx) from both numerator and denominator, you will reach at,
(1+sinx.cosx)÷(sinx.cosx)
And it is equal to secx.cosecx + 1
If u dont understand any step ask me….
Here it is…….
Hope it helps………