Question

sec theta -tan theta/sec theta+tan theta=cos^2theta/(1+sin theta)^2​

Answer 1

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Answer 2

$\LARGE{\bf{\underline{\underline{TO \ FIND:-}}}}$

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

$\LARGE{\bf{\underline{\underline{SOLUTION:-}}}}$

SIMPLIFY LHS AND RHS SEPARATELY.

RHS:

$\sf \to \dfrac{cos^2 \theta}{(1+sin \theta)^2}$

Use sin²θ + cos²θ = 1 ⇒ cos²θ = 1 - sin²θ

$\sf \to \dfrac{(1-sin^2 \theta)}{(1+sin \theta)^2}$

Use a² - b² = (a + b) (a - b) and (a + b)² = (a + b)(a + b)

$\sf \to \dfrac{(1-sin \theta)\cancel{(1+sin \theta)}}{(1+sin \theta)\cancel{(1+sin \theta)}}$

$\sf \to \boxed{\sf{\green{\dfrac{(1-sin \theta)}{(1+sin \theta)}}}}$

LHS:

$\to \ \sf \dfrac{sec \theta - tan \theta}{sec \theta + tan \theta}$

Use secθ = 1/cosθ and tanθ = sinθ/cosθ.

$\to \ \sf \dfrac{\dfrac{1}{cos \theta} - \dfrac{sin \theta}{cos \theta} }{\dfrac{1}{cos \theta} + \dfrac{sin \theta}{cos \theta} }$

Simplify.

$\to \ \sf \dfrac{\dfrac{1 - sin \theta}{cos \theta} }{ \dfrac{1+sin \theta}{cos \theta} }$

$\to \ \sf \dfrac{1 - sin \theta}{cos \theta} } \times \dfrac{cos \theta}{1 + sin \theta}$

$\sf \to \boxed{\sf{\green{\dfrac{(1-sin \theta)}{(1+sin \theta)}}}}$

LHS = RHS.

HENCE PROVED.

FUNDAMENTAL TRIGONOMETRIC IDENTITIES:

$\boxed{\substack{\displaystyle \sf sin^2 \theta+cos^2 \theta = 1 \\\\ \displaystyle \sf 1+cot^2 \theta=cosec^2 \theta \\\\ \displaystyle \sf 1+tan^2 \theta=sec^2 \theta}}$

T-RATIOS:

[tex]\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3} }{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }& 1 & \sqrt{3} & \rm Not \: De fined \\ \\ \rm cosec A & \rm Not \: De fined & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm Not \: De fined \\ \\ \rm cot A & \rm Not \: De fined & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0 \end{array} [/tex]

Regards,

SujalSirmilla

Ex-brainly star.