Question

Prove: secA(1−sinA)(secA+tanA)=1
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Answer 1

ANSWER:

To Prove:

• sec A(1 - sin A)(sec A + tan A) = 1

Proof:

We need to prove that,

$\implies\sec A(1-\sin A)(\sec A+\tan A)=1$

Taking LHS,

$\implies\sec A(1-\sin A)(\sec A+\tan A)$

Multiplying sec A with the first bracket,

$\implies\sec A(1-\sin A)(\sec A+\tan A)$

$\implies(\sec A-\sec A\sin A)(\sec A+\tan A)$

We know that,

$\hookrightarrow\sec\theta=\dfrac{1}{\cos\theta}$

So,

$\implies(\sec A-\sec A\sin A)(\sec A+\tan A)$

$\implies\bigg(\sec A-\dfrac{1}{\cos A}\cdot\sin A\bigg)\bigg(\sec A+\tan A\bigg)$

$\implies\bigg(\sec A-\dfrac{\sin A}{\cos A}\bigg)\bigg(\sec A+\tan A\bigg)$

We know that,

$\hookrightarrow\dfrac{\sin\theta}{\cos\theta}=\tan\theta$

So,

$\implies\bigg(\sec A-\dfrac{\sin A}{\cos A}\bigg)\bigg(\sec A+\tan A\bigg)$

$\implies\left(\sec A-\tan A\right)\left(\sec A+\tan A\right)$

We know that,

$\hookrightarrow (x-y)(x+y)=x^2-y^2$

So,

$\implies\left(\sec A-\tan A\right)\left(\sec A+\tan A\right)$

$\implies\sec^2A-\tan^2A$

We know that,

$\hookrightarrow\sec^2\theta-\tan^2\theta=1$

So,

$\implies\sec^2A-\tan^2A$

$\implies\bf 1 = RHS$

As, LHS = RHS,

HENCE PROVED!!

Answer 2

Step-by-step explanation:

$we \: have \: to \: prove \\ \: \: secA(1 - sinA) \: (secA + tanA) = 1 \\ proof: \\ let \: us \: start \: with \: \: l.h.s. \\ = (secA - sinA \times secA)(secA + tanA) \\ = (secA - tanA)(secA + tanA) \: (sec A = \frac{1}{cosA} \: and \: \frac{sinA}{cosA} =tanA \\ \\ = ({sec}^{2} A - {tan}^{2} A) \: \: \: as \: (a + b)(a - b) = {a}^{2} - {b}^{2} \\ = {sec}^{2} A - {tan}^{2} A \\ = 1. \\ = r.h.s. \\ \\ hence \: proved.$

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