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Prove: (TanA + CosecB)^2 - (CotB - SecA)^2 = 2TanA.CotB(CosecA + SecB)?
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Prove: (TanA + CosecB)^2 - (CotB - SecA)^2 = 2TanA.CotB(CosecA + SecB)...
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RHS=2tanA.cotB(cosecA+secB)

=(2tanA.cotB.cosecA )+(2tanA.cotB.secB)

=2[(sinA/cosA).(cosB/sinB).(1/sinA)+(sinA/cosA).(cosB.sinB).(1/cosB)]

=2(secA.cotB + tanA.cosecB)






LHS==== (tanA+cosecB)^2-(cotB-secA)^2

=tan^2A+cosec^2B+2.tanA.cosecB-(cot^2B+sec^2A-2.cotB.secA)

=tan^2A + cosec^2B +2tanA.cosecB-cot^B-sec^A+cotB.secA

=cosec^2B-cot^2B + tan^2-sec^2A + 2cotB.secA +2 tanA.cosecB

=1-1+2(cotB.secA + tanA .cosecB)

=2(cotB.secA + tanA.cosecB)






therefore,
.....LHS=RHS

hence proved
Community Answer
Prove: (TanA + CosecB)^2 - (CotB - SecA)^2 = 2TanA.CotB(CosecA + SecB)...
Proof:

We will start by simplifying each term of the given equation.

Expanding the first term:
(TanA * CosecB)^2
= (SinA/CosA * 1/SinB)^2 (using the identity CosecB = 1/SinB and TanA = SinA/CosA)
= (SinA/CosA)^2 * (1/SinB)^2
= Sin^2A/Cos^2A * 1/Sin^2B
= (1/Cos^2A) * (1/Sin^2B) * Sin^2A
= Sin^2A/(Sin^2B * Cos^2A)

Expanding the second term:
(CotB - SecA)^2
= (CosB/SinB - 1/CosA)^2 (using the identity CotB = CosB/SinB and SecA = 1/CosA)
= (Cos^2B/Sin^2B) - 2(CosB/SinB)(1/CosA) + (1/Cos^2A)
= (Cos^2B/Sin^2B) - 2(CosB/CosA)(1/SinB) + (1/Cos^2A)
= (Cos^2B/Cos^2A * Sin^2A/Sin^2B) - 2(SinA/CosA * CosB/SinB)(1/SinB) + (1/Cos^2A)
= Sin^2A/(Sin^2B * Cos^2A) - 2CosB * TanA/SinB + 1/Cos^2A

Expanding the third term:
2TanA.CotB(CosecA + SecB)
= 2(SinA/CosA)(CosB/SinB)(1/SinA + 1/CosA)
= 2(SinA/CosA)(CosB/SinB)(CosA/SinA + SinB/CosB)
= 2SinB + 2TanA * CosB/SinB

Substituting the simplified terms into the equation:
(TanA*CosecB)^2 - (CotB - SecA)^2 = 2TanA.CotB(CosecA + SecB)
Sin^2A/(Sin^2B * Cos^2A) - (Sin^2A/(Sin^2B * Cos^2A) - 2CosB * TanA/SinB + 1/Cos^2A) = 2SinB + 2TanA * CosB/SinB
2CosB * TanA/SinB - 1/Cos^2A = 2SinB + 2TanA * CosB/SinB
2CosB * TanA/SinB - 2TanA * CosB/SinB = 2SinB + 1/Cos^2A
2TanA * CosB/SinB(1 - Cos^2B) = 2SinB + 1/Cos^2A
2TanA * Sin^2B/SinB * 1/S
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Prove: (TanA + CosecB)^2 - (CotB - SecA)^2 = 2TanA.CotB(CosecA + SecB)?
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Prove: (TanA + CosecB)^2 - (CotB - SecA)^2 = 2TanA.CotB(CosecA + SecB)? for Class 10 2024 is part of Class 10 preparation. The Question and answers have been prepared according to the Class 10 exam syllabus. Information about Prove: (TanA + CosecB)^2 - (CotB - SecA)^2 = 2TanA.CotB(CosecA + SecB)? covers all topics & solutions for Class 10 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Prove: (TanA + CosecB)^2 - (CotB - SecA)^2 = 2TanA.CotB(CosecA + SecB)?.
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