Question

Plzz solve these....
(iii) and (iv) ​

Answer 1

Answer:

Given:

(III) The equation is,

1-sinA/1+sinA=(secA-tanA)²

Take LHS,

1-sinA/1+sinA

→On rationalizing,

=1-sinA/1+sinA×1-sinA/1-SinA

=(1-sinA)²/1-sin²A

=1+sin²A-2sinA/cos²A

=Sec²A+tan²A-2secA.tanA

=(secA-tanA)²

(IV)

AS we know that,

Sec²x-tan²x=1

sin²x+cos²x=1

→The given equation is,

(secA+cosA)(secA-cosA)

= Sec²A-cos²A

= 1+tan²A-(1-sin²A)

= 1+tan²A-1+sin²A

= tan²A+sin²A

Answer 2

Given:

$(1) \: \dfrac{1 - \sin\theta}{1 + \sin\theta} = (\sec\theta - \tan\theta)^{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ (2) \: (\sec\theta + \cos\theta)(\sec\theta \: - \cos\theta) = {\tan}^{2}\theta + {\sin}^{2}\theta$

To Prove:

LHS = RHS

Answer:

(1):

First we will solve LHS.

LHS:

Rationalising the denominator, we get:

$\dfrac{1 - \sin\theta}{1 + \sin\theta} \times \dfrac{1 -\sin\theta}{1 - \sin\theta} \: \: \: \: \: \: \\ \\ (a - b)(a + b) = {a}^{2} - {b}^{2} \\ \\ (a - b)^{2} = {a}^{2} + {b}^{2} - 2ab \\ \\ \frac{(1 - \sin\theta)^{2} }{ {1}^{2} - {\sin}^{2} \theta} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \frac{1 + {\sin}^{2}\theta - 2\sin\theta }{1 - {\sin}^{2}\theta } \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ 1 - {\sin}^{2} \theta \: = \: {\cos}^{2} \theta \: \: \: \: \: \: \: \: \: \: \: \:$

$\dfrac{1 + {\sin}^{2}\theta - 2\sin\theta }{ {\cos}^{2}\theta } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \frac{1}{\cos^{2}\theta } + \frac{ {\sin}^{2}\theta }{ {\cos}^{2}\theta } - \frac{2\sin\theta}{\cos\theta \times \cos\theta}$

We can write:

$\\ \\ \dfrac{1}{ {\cos}^{2} \theta} = {\sec}^{2} \theta \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \dfrac{ {\sin}^{2} \theta}{ {\cos}^{2}\theta } = {\tan}^{2} \theta \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ {\sec}^{2} \theta + {\tan}^{2} \theta - 2 \: \dfrac{\sin\theta}{\cos\theta} \times \sec\theta \\ \\ {\sec}^{2} \theta + {\tan}^{2} \theta - 2\sec\theta\tan\theta \: \: \: \: \: \\ \\ (\sec\theta - \tan\theta)^{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

LHS = RHS .

Hence Proved.

(2):

LHS:

$\left( \dfrac{1}{\cos\theta} + \cos\theta\right)\left( \dfrac{1}{\cos\theta} - \cos\theta \right)$

Taking LCM , we get:

$\\ \\ \left( \dfrac{1 + {\cos}^{2}\theta }{\cos\theta} \right)\left( \dfrac{1 - {\cos}^{2} \theta}{\cos\theta} \right) \\ \\ 1 - {\cos}^{2} \theta = {\sin}^{2} \theta \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \left( \dfrac{\left(1 + {\cos}^{2}\theta\right) {\sin}^{2}\theta }{ {\cos}^{2} \theta} \right) \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \dfrac{ {\sin}^{2}\theta + {\sin}^{2}\theta {\cos}^{2}\theta }{ {\cos}^{2}\theta } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \dfrac{ {\sin}^{2}\theta }{ {\cos\theta}^{2} } + \frac{ {\sin}^{2}\theta \: {\cos}^{2}\theta }{ {\cos}^{2} \theta} \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ {\tan}^{2} \theta + {\sin}^{2} \theta \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

LHS = RHS

Hence Proved.