Question

Simplify the trigonomic expression 1/1+ sin

Answer 1

Answer:

$\sec^2\theta-\tan\theta\sec\theta$

Step-by-step explanation:

We have to simplify the trigonometric expression $\frac{1}{1+\sin\theta}$

Let us multiply the numerator and denominator by $1-\sin\theta$

$\frac{1}{1+\sin\theta}\times\frac{1-\sin\theta}{1-\sin\theta}$

In the denominator apply the formula for difference of squares

$a^2-b^2=(a+b)(a-b)$

Thus, the denominator will become

$(1+\sin\theta)(1-\sin\theta)=1-\sin^2\theta$

Thus, the expression is

$\frac{1-\sin\theta}{1-\sin^2\theta}$

Using the relation $1-\sin^2\theta=\cos^2\theta$

$\frac{1-\sin\theta}{\cos^2\theta}$

We can rewrite this expression as

$\frac{1}{\cos^2\theta}-\frac{\sin\theta}{\cos^2\theta}$

Finally, the simplified expression is

$\sec^2\theta-\tan\theta\sec\theta$

Answer 2

Answer:

sec²θ - tanθ(secθ) is the simplification of given trigonometric expression.

Step-by-step explanation:

We have given the following  trigonometric expression:

1/1+sinθ

We have to simplify this   trigonometric expression.

For this, we multiply and divide the given  trigonometric expression by

1-sinθ.

(1/1+sinθ ) ÷ ( 1-sinθ)/(1-sinθ)

1(1-sinθ) / (1-sinθ)(1+sinθ)

(1-sinθ) / (1-sin²θ)

We know that 1-sin²θ = cos²θ.

We replace the denominator of above trigonometric expression by  cos²θ.

( 1-sinθ) / cos²θ

(1/cos²θ) - (sinθ/cos²θ)

(1/cos²θ) - ( sinθ/cosθ)(1/cosθ)

As we know that 1/cos²θ = sec²θ, sinθ/cosθ = tanθ and 1/cosθ = secθ.So we have,

sec²θ - tanθ(secθ) is the simplification of given trigonometric expression.