The correct option is B secA tanA 1/secA+tanA=1/secA+tanA×secA tanA/secA tanA =secA tanA/sec^2A tan^2A = secA tanA [∵ sec^2A tan^2A=1] ...

1/ A+tan A=?

Question

$\frac{1}{secA+tanA}$ =?

secA+tanA

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secA-tanA

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$s e c^{2} A$

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Solution

The correct option is B
secA-tanA

$\frac{1}{secA+tanA}$ = $\frac{1}{secA+tanA} \times \frac{secA-tanA}{secA-tanA}$

= $\frac{secA-tanA}{s e c^{2} A - t a n^{2} A}$

= secA-tanA [ $∵ s e c^{2} A - t a n^{2} A = 1$ ]

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Similar questions

\frac{tan A + sec A - 1}{tan A - sec A + 1} = tan A + sec A

1+tanA*tanA/1+cotA*cotA=1-tanA/1-cotA

Prove it

\frac{tan A}{1 - cot A} + \frac{cot A}{1 - tan A} = 1 + tan A + cot A

\frac{tan A}{1 - cot A} + \frac{cos A}{1 - tan A} = 1 + tan A + cot A = 1 + sec A csc A

\frac{tan A}{1 - cot A} + \frac{cot A}{1 - tan A} = sec A c o s e c A + 1