Mohan is a ten year poor boy. He is pushed to a firework factory. He is some time so hungry that he eat glue that is use to make fireworks and match box. The people running the factory knew about it and started adding poison to the glue to ensure that the other children do not eat the same? Help Mohan to give a safe environment. how will you rescue Mohan
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 ther 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.