Tyra has 12 2/3 cups of sugar. If r is the amount of sugar called for in one batch of cookies,
the expression 12 2/3 ÷ r can be used to find how many batches Tyra can make with the sugar she has.
If r is 2/3 cup, how many batches of cookies can Tyra make?
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______batches
Answers
Answer:
\LARGE{\bf{\underline{\underline{GIVEN:-}}}}GIVEN:−
\sf \bullet \ \ \dfrac{(1+sinA-cosA)^2}{(1+sinA+cosA)^2}∙ (1+sinA+cosA)2(1+sinA−cosA)2
\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 .
\sf \to \dfrac{(cos^2-2sincos+sin^2-2cos+2sin+1)}{(cos^2+2sincos+sin^2+2cos+2sin+1)}→(cos2+2sincos+sin2+2cos+2sin+1)(cos2−2sincos+sin2−2cos+2sin+1)
Rearrange the terms.
\sf \to \dfrac{(cos^2+sin^2-2sincos-2cos+2sin+1)}{(cos^2+sin^2+2sincos+2cos+2sin+1)}→(cos2+sin2+2sincos+2cos+2sin+1)(cos2+sin2−2sincos−2cos+2sin+1)
We know that cos²A+sin²A=1.
\sf \to \dfrac{1-2sincos-2cos}{2sin+1}→2sin+11−2sincos−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+11−2cos(sin+2)
Now, rearrange the terms, add 1 and 1 and take 2 common.
\to\sf\dfrac{1+1+2sin-2cos}{sin+1}→sin+11+1+2sin−2cos
\to\sf\dfrac{2+2sin-2cos}{sin+1}→sin+12+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+cosA1−cosA
LHS=RHS.
HENCE PROVED!
FUNDAMENTAL TRIGONOMETRIC RATIOS:
\begin{gathered} \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}\end{gathered}sin2θ+cos2θ=11+cot2θ=cosec2θ1+tan2θ=sec2θ
T-RATIOS: