Question

Make a precis of this


Games though essential, should not become the be all of students life. One must never

devote more than an hour to sports and after that concentrate on studies. If games are

played without the spirit of sportsmanship , It can lead to bad blood and quarrels.Some

players when loosing the game or match intentionally play foul to interrupt the match .

But in spite all these minor defects , sports are very useful in keeping students busy and in developing their personalities . India expects its citizen to have qualities of true

sportsmanship. If we all acquire these qualities ,there will be no narrow - mindedness ,

no corruption and no injustice. There will be independence in all the real sense.​

Answer 1

Answer:

The functional substance, or parenchyma, of the kidney is divided into two major structures: the outer renal cortex and the inner renal medulla. Grossly, these structures take the shape of eight to 18 cone-shaped renal lobes, each containing renal cortex surrounding a portion of medulla called a renal pyramid

Answer 2

Explanation:

\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