Seven people P, Q, R, S, T, U and V are sitting in a circle facing outwards. Each of them belongs to – Economist, Minister, Engineer and Mathematician but not necessary in the same order. There are four Engineers. R sits third to the left of S. T is neither an immediate neighbour of S nor of R. The one sitting exactly between S and U is a Mathematician. V sits third to the left of P and V is a Minister. Q is an Engineer. One of Q’s neighbours is a Economist. Economist is not sitting between Engineer and Minister.
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)}→
(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
Explanation:
It Can be defined as
Momentum, product of the mass of a particle and its velocity. Momentum is a vector quantity; i.e., it has both magnitude and direction. Isaac Newton's second law of motion states that the time rate of change of momentum is equal to the force acting on the particle. See Newton's laws of motion. Momentum.