Question

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

Answer 1

Answer:

Newton's law of universal gravitation is usually stated as that every particle attracts every other particle in the universe with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centersNewton's law of universal gravitation is usually stated as that every particle attracts every other particle in the universe with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers\huge{\fcolorbox{yellow}{red}{\large{\fcolorbox{blue}{pink}{{\fcolorbox{orange}{aqua}{ ANSWER❤ }}}}}}

ANSWER❤

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ANSWER❤

Newton's law of universal gravitation is usually stated as that every particle attracts every other particle in the universe with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers

Answer 2

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

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

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

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:

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