Question

1-sin e 1+ sin e seco - tano​

Answer 1

Answer:

To prove that: \sqrt{\dfrac{1-\sin \theta}{1+\sin \theta} } =\sec \theta-\tan \theta.

1+sinθ

1−sinθ

=secθ−tanθ.

Solution:

L.H.S. = \sqrt{\dfrac{1-\sin \theta}{1+\sin \theta} }

1+sinθ

1−sinθ

To rationalizing denominator, we get

= \sqrt{\dfrac{1-\sin \theta}{1+\sin \theta} }

1+sinθ

1−sinθ

× \sqrt{\dfrac{1-\sin \theta}{1-\sin \theta} }

1−sinθ

1−sinθ

Using the algebraic identity:

(a + b)(a - b) = a^{2}a

2

- b^{2}b

2

= \sqrt{\dfrac{(1-\sin \theta)^2}{1^2-\sin^2 \theta} }

1

2

−sin

2

θ

(1−sinθ)

2

= \dfrac{1-\sin \theta}{\sqrt{1-\sin^2 \theta}}

1−sin

2

θ

1−sinθ

Using the trigonometric identity:

\sin^2 Asin

2

A + \cos^2 Acos

2

A = 1

⇒ \cos^2 Acos

2

A = 1 - \sin^2 Asin

2

A

= \dfrac{1-\sin \theta}{\sqrt{\cos^2 \theta} }

cos

2

θ

1−sinθ

= \dfrac{1-\sin \theta}{\cos \theta}

cosθ

1−sinθ

= \dfrac{1}{\cos \theta}-\dfrac{\sin \theta}{\cos \theta}

cosθ

1

−

cosθ

sinθ

= \sec \theta-\tan \thetasecθ−tanθ

= R.H.S., proved.

Thus, \sqrt{\dfrac{1-\sin \theta}{1+\sin \theta} } =\sec \theta-\tan \theta

1+sinθ

1−sinθ

=secθ−tanθ , proved.