At the brilliant stars.
A. Onomatopoeia
B. Transferred Epithet
C. Tautology
D. Oxymoron
6
Answer the questions in five to six sentences each :
17. What does the young man mean by “great honey-coloured / Ramparts at your
ear" ?
18. “Belinda tickled him, she tickled him unmerciful Why?
19. Explain : “It sits looking/over harbour and city / on silent haunches
OR
19. How does the crow change the mood of Robert Frost ?
Answer the questions in five to six sentences each:
20. How did the necklace change the life of Matilda Loisel ?
21. How does Bholi's teacher play an important role in changing the course of
her life?
22. Why does not Anil hand the thief over to the police ?
OR
22. Horace Danby was a meticulous planner but still he faltered. Where did he
go wrong and why?
Answers
Scope of public finance includes : ... Which one of the following is not a method for redeeming public debt? ... When the government raises revenue by borrowing from within the country isCalculate the concentration of hydrogen (H+) ions by dividing the molecules of hydrogen ions by the volume, in liters, of the solution. Take the negative log of this number. The result should be between zero and 14, and this is the pH.
good evening
Explanation:
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)}→
(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.
\to\sf\dfrac{1+1+2sin-2cos}{sin+1}→
sin+1
1+1+2sin−2cos
\to\sf\dfrac{2+2sin-2cos}{sin+1}→
sin+1
2+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+cosA
1−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}
sin
2
θ+cos
2
θ=1
1+cot
2
θ=cosec
2
θ
1+tan
2
θ=sec
2
θ
T-RATIOS:
\begin{gathered}\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}\end{gathered}
∠A
sinA
cosA
tanA
cosecA
secA
cotA
0
∘
0
1
0
NotDefined
1
NotDefined
30
∘
2
1
2
3
3
1
2
3
2
3
45
∘
2
1
2
1
1
2
2
1
60
∘
2
3
2
1
3
3
2
2
3
1
90
∘
1
0
NotDefined
1
NotDefined
0