Question

Sec theta-1/sec theta +1=(sin theta/1+cos theta)2

Answer 1

Hey there!

L.H.S =(secA – 1)/ (sec A + 1)

(1/cosA - 1)/ (1/cos A +1)

(1 – cosA/cosA)/ (1 + cosA/cosA)

(1 – cosA)/ (1 + cosA)

Multiplying numerator and denominator by (1 + cosA), we get

(1 - cosA)(1 + cosA)/ (1 + cosA)²

(1 – cos²A)/ (1 + cosA)²

sin²A/ (1 + cosA)²

[ sinA/(1 + cosA) ]² = R.H.S

Cheers!

Answer 2

GIVEN:

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

Or using $\sf \left(\dfrac{x}{y} \right)^2=\dfrac{x^2}{y^2}$, we can write the above identity as:

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

SOLUTION:

Simplify LHS and RHS separately.

LHS:

$\sf \bullet \ \dfrac{sec \theta - 1}{sec \theta +1}$

We know that secθ=1/cosθ.

$\sf \bullet \ \dfrac{\dfrac{1}{cos \theta} - 1}{\dfrac{1}{cos \theta} +1}$

$\sf \bullet \ \dfrac{\dfrac{1-cos \theta}{cos \theta} }{\dfrac{1+ cos \theta}{cos \theta} }$

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

$\sf \bullet \ \boxed{\sf{\green{\dfrac{1-cos \theta}{1+cos \theta}}}}$

RHS:

$\sf \bullet \ \left(\dfrac{sin^2 \theta}{(1+cos \theta)^2 } \right)$

We know that sin²θ = 1 - cos²θ.

$\sf \bullet \ \left(\dfrac{1-cos ^2 \theta}{(1+cos \theta)^2 } \right)$

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

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

$\sf \bullet \ \boxed{\sf{\green{\dfrac{1-cos \theta}{1+cos \theta}}}}$

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

Sujal Sirimilla

Ex brainly star.