If A+B=90∘ then prove sinAsecBtanAtanB+ tanAcotB−cos2Asin2B=tanA

Given that, A+B=90∘⟹B=90∘−AtanB=tan(90∘−A)=cotAsecB=sec(90∘−A)=cscAsinB=sin(90∘−A)=cosA∴secAsecBtanAtanB+tanBtanA−cos2Asin2B=sinAcscAtanAcotA+cotAtanA−cos2Acos2A=11+1−1=2−1=1=1

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If A+ B =90^{} then prove {{ A B + A B}{ A B} -{^{2}B}{^{2

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If $A + B = 9 0^{\circ}$ then prove $\frac{tan A tan B + tan A cot B}{sin A sec B} - \frac{sin ^{2} B}{cos ^{2} A} = tan A$

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Elijah Discussed

A + B = 9 0^{\circ}

then prove

\frac{tan A tan B + tan A cot B}{sin A sec B} - \frac{sin ^{2} B}{cos ^{2} A} = tan A

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Text solutionVerified

Given that,

A + B = 9 0^{\circ}

⟹ B = 9 0^{\circ} - A

tan B = tan (9 0^{\circ} - A) = cot A

sec B = sec (9 0^{\circ} - A) = csc A

sin B = sin (9 0^{\circ} - A) = cos A

∴ \frac{tan A tan B + tan B tan A}{sec A sec B} - \frac{sin ^{2} B}{cos ^{2} A}

= \frac{tan A cot A + cot A tan A}{sin A csc A} - \frac{cos ^{2} A}{cos ^{2} A}

= \frac{1 + 1}{1} - 1

= 2 - 1 = 1 = 1

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Question Text	If $A + B = 9 0^{\circ}$ then prove $\frac{tan A tan B + tan A cot B}{sin A sec B} - \frac{sin ^{2} B}{cos ^{2} A} = tan A$
Answer Type	Text solution:1
Upvotes	150

If A+B=90∘ then prove sinAsecBtanAtanB+ tanAcotB​−cos2Asin2B​​=tanA

Text solutionVerified

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If $A + B = 9 0^{\circ}$ then prove $\frac{tan A tan B + tan A cot B}{sin A sec B} - \frac{sin ^{2} B}{cos ^{2} A} = tan A$