What is the area of a syude WUSC SIU 13 mm.
5. Length and breadth of a bulletin board are r сm and t cm, respectively.
(a)What will be the length (in cm) of the aluminium strip required to frame the board, if 10 cm extra strip
is required to fix it properly.
(b) Ifr nails are used to repair one board, how many nails will be required to repair 15 such boards?
(c) If 500 sq. cm extra cloth per board is required to cover the edges, what will be the total area of the cloth
required to cover 8 such boards?
(d) What will be the expenditure for making 23 boards, if the carpenter charges x per board?
Answers
Answer:
Answer:
\LARGE{\bf{\underline{\underline{GIVEN:-}}}}
GIVEN:−
\sf \bullet \ \ \dfrac{(1+sinA-cosA)^2}{(1+sinA+cosA)^2} < /p > < p >∙
(1+sinA+cosA)
2
(1+sinA−cosA)
2
</p><p>
\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)}→
(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.
Answer:
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