Question

solve this question​

Answer 1

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

$\sf \bullet \ \ \dfrac{(1+sinA-cosA)^2}{(1+sinA+cosA)^2} = \dfrac{1-cos}{1+cos}$

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

LHS:

$\sf \to \dfrac{(1+sinA-cosA)^2}{(1+sinA+cosA)^2}$

Expand the fractions using (a+b+c)²=a²+b²+c²+2ab+2bc+2ca.

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

Rearrange the terms.

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

We know that cos²A+sin²A=1.

$\sf \to \dfrac{1-2sincos-2cos}{2sin+1}$

Now here, take -2cos common from the numerator and +2cos common from the denominator.

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

Now, rearrange the terms, add 1 and 1 and take 2 common.

$\to\sf\dfrac{1+1+2sin-2cos}{sin+1}$

$\to\sf\dfrac{2+2sin-2cos}{sin+1}$

Take 2 common.

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

Take (1+sin) common.

$\to \sf \dfrac{ \not{2}\cancel{(1+sin)}(1 - cos) }{\not{2}\cancel{(1+sin )}(1 + cos )}$

$\to \sf{\red{\dfrac{1-cosA}{1+cosA} }}$

LHS=RHS.

HENCE PROVED!

FUNDAMENTAL TRIGONOMETRIC RATIOS:

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

T-RATIOS:

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

Answer 2

Answer:

GIVEN:−

\sf \bullet \ \ \dfrac{(1+sinA-cosA)^2}{(1+sinA+cosA)^2} = \dfrac{1-cos}{1+cos}∙

(1+sinA+cosA)

2

(1+sinA−cosA)

2

=

1+cos

1−cos

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

SOLUTION:−

LHS:

\sf \to \dfrac{(1+sinA-cosA)^2}{(1+sinA+cosA)^2}→

(1+sinA+cosA)

2

(1+sinA−cosA)

2

Expand the fractions using (a+b+c)²=a²+b²+c²+2ab+2bc+2ca.

\sf \to \dfrac{(cos^2-2sincos+sin^2-2cos+2sin+1)}{(cos^2+2sincos+sin^2+2cos+2sin+1)}→

(cos

2

+2sincos+sin

2

+2cos+2sin+1)

(cos

2

−2sincos+sin

2

−2cos+2sin+1)

Rearrange the terms.

\sf \to \dfrac{(cos^2+sin^2-2sincos-2cos+2sin+1)}{(cos^2+sin^2+2sincos+2cos+2sin+1)}→

(cos

2

+sin

2

+2sincos+2cos+2sin+1)

(cos

2

+sin

2

−2sincos−2cos+2sin+1)

We know that cos²A+sin²A=1.

\sf \to \dfrac{1-2sincos-2cos}{2sin+1}→

2sin+1

1−2sincos−2cos

Now here, take -2cos common from the numerator and +2cos common from the denominator.

\sf \to \dfrac{1-2cos(sin+2)}{2sin+1}→

2sin+1

1−2cos(sin+2)

Now, rearrange the terms, add 1 and 1 and take 2 common.

\to\sf\dfrac{1+1+2sin-2cos}{sin+1}→

sin+1

1+1+2sin−2cos

\to\sf\dfrac{2+2sin-2cos}{sin+1}→

sin+1

2+2sin−2cos

Take 2 common.

\to \sf \dfrac{ 2(1+sin) -2cos(sin+1) }{ 2(1+sin) + 2cos(sin +1 ) }→

2(1+sin)+2cos(sin+1)

2(1+sin)−2cos(sin+1)

Take (1+sin) common.

\to \sf \dfrac{ \not{2}\cancel{(1+sin)}(1 - cos) }{\not{2}\cancel{(1+sin )}(1 + cos )}→



2

(1+sin)

(1+cos)



2

(1+sin)

(1−cos)

\to \sf{\red{\dfrac{1-cosA}{1+cosA} }}→

1+cosA

1−cosA

LHS=RHS.

HENCE PROVED!

FUNDAMENTAL TRIGONOMETRIC RATIOS:

\begin{gathered}\begin{gathered}\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}}\end{gathered}\end{gathered}

sin

2

θ+cos

2

θ=1

1+cot

2

θ=cosec

2

θ

1+tan

2

θ=sec

2

θ

T-RATIOS:

\begin{gathered}\begin{gathered}\boxed{\boxed{\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}}}\end{gathered}\end{gathered}

∠A

sinA

cosA

tanA

cosecA

secA

cotA

0

∘

0

1

0

NotDefined

1

NotDefined

30

∘

2

1

2

3

3

1

2

3

2

3

45

∘

2

1

2

1

1

2

2

1

60

∘

2

3

2

1

3

3

2

2

3

1

90

∘

1

0

NotDefined

1

NotDefined

0