Question

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

Answer 1

Step-by-step explanation:

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

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

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

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.

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

$\to \sf{\red{\dfrac{1-cosA}{1+cosA} }}→ </p><p>$

LHS=RHS.

HENCE PROVED!

FUNDAMENTAL TRIGONOMETRIC RATIOS:

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

T-RATIOS:

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

Answer 2

Answer:

Basic Trigonometric Identities for Sine and Cos

cos

2

(

A

)

+

sin

2

(

A

)

=

1

If A + B = 180° then:

sin

(

A

)

=

sin

(

B

)

cos

(

A

)

=

−

cos

(

B

)

If A + B = 90° then:

sin

(

A

)

=

cos

(

B

)

cos

(

A

)

=

sin

(

B

)

Half-angle formulas

sin

(

A

2

)

=

√

±

1

−

cos

(

A

)

2

If

A

2

lies in quadrant I or II

If

A

2

lies in quadrant III or IV

cos

(

A

2

)

=

√

±

1

+

cos

(

A

)

2

If

A

2

lies in quadrant I or IV

If

A

2

lies in quadrant II or III

Double and Triple Angle Formulas

sin

2

A

=

2

sin

A

cos

A

cos

2

A

=

cos

2

A

–

sin

2

A

=

2

cos

2

–

1

=

1

−

sin

2

A

sin

3

A

=

3

sin

A

–

4

sin

3

A

cos

3

A

=

4

cos

3

A

–

3

cos

A

sin

4

A

=

4

cos

3

A

sin

A

–

4

cos

A

sin

3

A

cos

4

A

=

cos

4

A

–

6

cos

2

A

sin

2

A

+

sin

4

A

sin

2

A

=

1

–

cos

(

2

A

)

2

cos

2

A

=

1

+

cos

(

2

A

)

2

Sum and Difference of Angles

sin

(

A

+

B

)

=

sin

(

A

)

cos

(

B

)

+

cos

(

A

)

sin

(

B

)

sin

(

B

)

sin

(

A

−

B

)

=

sin

(

A

)

cos

(

B

)

−

cos

(

A

)

sin

(

B

)

cos

(

A

+

B

)

=

cos

(

A

)

cos

(

B

)

−

sin

(

A

)

sin

(

B

)

cos

(

A

−

B

)

=

cos

(

A

)

cos

(

B

)

+

sin

(

A

)

sin

(

B

)

sin

(

A

+

B

+

C

)

=

sin

A

cos

B

cos

C

+

cos

A

sin

B

cos

C

+

cos

A

cos

B

sin

C

−

sin

A

sin

B

sin

C

cos

(

A

+

B

+

C

)

=

cos

A

cos

B

cos

C

−

sin

A

sin

B

cos

C

−

sin

A

cos

B

sin

C

−

sin

A

cos

B

sin

C

−

cos

A

sin

B

sin

C

sin

A

+

sin

B

=

2

sin

(

A

+

B

)

2

cos

(

A

−

B

)

2

sin

A

–

sin

B

=

2

sin

(

A

−

B

)

2

cos

(

A

+

B

)

2

cos

A

+

cos

B

=

2

cos

(

A

+

B

)

2

(

A

+

B

)

2

cos

(

A

−

B

)

2

cos

A

+

cos

B

=

−

2

sin

(

A

+

B

)

2

sin

(

A

−

B

)

2

Product Identities

sin

(

x

)

cos

(

y

)

=

1

2

[

sin

(

x

+

y

)

+

sin

(

x

–

y

)

]

cos

(

x

)

sin

(

y

)

=

1

2

[

sin

(

x

+

y

)

–

sin

(

x

–

y

)

]

cos

(

x

)

cos

(

y

)

=

1

2

[

cos

(

x

–

y

)

+

cos

(

x

+

y

)

]

sin

(

x

)

sin

(

y

)

=

1

2

[

cos

(

x

–

y

)

–

cos

(

x

+

y

)

]