Question

$\huge\blue {\underline { \overline{ \bf \mid\: \: \: \: \: QUESTION \: \: \: \: \mid}}}$


(cos A × cosec B - sin A × sec B) / (cos A + sin A) = cosec A - sec A
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Answer 1

Answer:

Equality with sine, cosine, or tangent in them is called trigonometric equality. These are solved by some interrelations known beforehand. All the interrelations which relate sine, cosine, tangent, secant, cotangent, cosecant are called trigonometric identities. These trigonometric identities solve the equation and make them simpler to understand for proof. These are the main and crucial steps to take us nearer to the result.

Let us consider our question.

cosec A + sec A = cosec B + sec B

By general trigonometric knowledge, we know that

cosecA=1sinA;secA=1cosAcosec⁡A=1sin⁡A;sec⁡A=1cos⁡A

By substituting these values into our original equation, we get,

1sinA+1cosA=1sinB+1cosB1sin⁡A+1cos⁡A=1sin⁡B+1cos⁡B

By subtracting 1sinB1sin⁡B on both the sides, we get the equation as,

1sinA−1sinB+1cosA=1cosB1sin⁡A−1sin⁡B+1cos⁡A=1cos⁡B

By subtracting 1cosA1cos⁡A on both the sides, we get the equation as,

1sinA−1sinB=1cosB−1cosA1sin⁡A−1sin⁡B=1cos⁡B−1cos⁡A

By taking the least common multiple on both the sides, we get,

sinB−sinAsinAsinB=cosA−cosBcosAcosBsin⁡B−sin⁡Asin⁡Asin⁡B=cos⁡A−cos⁡Bcos⁡Acos⁡B

By doing cross-multiplication, we get an equation of the form,

tanAtanB=sinB−sinAcosA−cosBtan⁡Atan⁡B=sin⁡B−sin⁡Acos⁡A−cos⁡B

By basic trigonometric knowledge, we get them as:

sinA−sinB=2sin(A−B2)cos(A+B2)sin⁡A−sin⁡B=2sin⁡(A−B2)cos⁡(A+B2)

cosA−cosB=−2sin(A+B2)sin(A−B2)cos⁡A−cos⁡B=−2sin⁡(A+B2)sin⁡(A−B2)

By taking “ – “ inside, we can write the formula as:

cosA−cosB=2sin(A+B2)sin(B−A2)cos⁡A−cos⁡B=2sin⁡(A+B2)sin⁡(B−A2)

By substituting these equations, into our equation, we get,

sinAsinBcosAcosB=2cos(A+B2)sin(B−A2)2sin(A+B2)sin(B−A2)sin⁡Asin⁡Bcos⁡Acos⁡B=2cos⁡(A+B2)sin⁡(B−A2)2sin⁡(A+B2)sin⁡(B−A2)

By canceling the common terms on the right-hand side, we get,

sinAcosA.sinBcosB=cos(A+B2)sin(A+B2)sin⁡Acos⁡A.sin⁡Bcos⁡B=cos⁡(A+B2)sin⁡(A+B2)

By basic knowledge of trigonometry, we know the relations as:

tanx=sinxcox;cotx=cosxsinxtan⁡x=sin⁡xcox;cot⁡x=cos⁡xsin⁡x

By substituting these into our equation, we get it as:

tanA.tanB=cot(A+B2)tan⁡A.tan⁡B=cot⁡(A+B2)

Hence proved.

Therefore, we have proved the required equation by the given condition.

Answer 2

it is right answer of ur Question