Question

sec A (1 - sin A) (sec A + tan A) = 1​

Answer 1

Solution :-

Here, L.H.S. = sec A (1 - sin A) (sec A + tan A)

R.H.S. = 1

Solving L.H.S. -

=> sec A (1 - sin A) (sec A + tan A)

=> $\frac{1}{cos\:\:A}(1 - sin A) (\frac{1}{cos\:\:A} + \frac{sin\:\:A}{cos\:\:A})$

[sec Ф = $\frac{1}{cos\:\:\theta}$  and tan Ф = $\frac{sin\:\:\theta}{cos\:\:\theta}$]

=> $\frac{1 - sin A}{cos\:\:A}\times \frac{1+sin\:\:A}{cos\:\:A}$

=> $\frac{(1 - sin A)(1+sin\:\:A)}{cos^{2} \:\:A}$

We know that,

(a-b)(a+b)= a² - b²

=> $\frac{(1^{2} - sin^{2} A)}{cos^{2} \:\:A}$

=> $\frac{(1 - sin^{2} A)}{cos^{2} \:\:A}$

[1 - sin² Ф = cos² Ф]

=> $\frac{cos^{2} \:\:A}{cos^{2} \:\:A}$

=> 1 = R.H.S.

More To Know :-

$\star$ cot θ = cos θ / sin θ

$\star$ cot θ = 1 / tan θ

$\star$ tan θ = sin θ / cos θ

$\star$ tan θ = 1 / cos θ

$\star$ sin² θ + cos² θ =1

$\star$ sec² θ - tan² θ =1

Answer 2

Solution :-

Here, L.H.S. = sec A (1 - sin A) (sec A + tan A)

R.H.S. = 1

Solving L.H.S. -

=> sec A (1 - sin A) (sec A + tan A)

=> $\frac{1}{cos\:\:A}(1 - sin A) (\frac{1}{cos\:\:A} + \frac{sin\:\:A}{cos\:\:A})$

[sec Ф = $\frac{1}{cos\:\:\theta}$  and tan Ф = $\frac{sin\:\:\theta}{cos\:\:\theta}$]

=> $\frac{1 - sin A}{cos\:\:A}\times \frac{1+sin\:\:A}{cos\:\:A}$

=> $\frac{(1 - sin A)(1+sin\:\:A)}{cos^{2} \:\:A}$

We know that,

(a-b)(a+b)= a² - b²

=> $\frac{(1^{2} - sin^{2} A)}{cos^{2} \:\:A}$

=> $\frac{(1 - sin^{2} A)}{cos^{2} \:\:A}$

[1 - sin² Ф = cos² Ф]

=> $\frac{cos^{2} \:\:A}{cos^{2} \:\:A}$

=> 1 = R.H.S.