Who was the “untutored genius” of India? (+1 class hornbill book chapter no. 4)
Answers
Answer:
Nek Chand, the director of The Rock gardens was the untutored genius who was the mastermind behind the creation of the Rock garden at Chandigarh which is his biggest contribution to outside art. The fiftieth issue of Raw Vision, a UK magazie featured "Women by the waterfall", one of his rock garden sculpture. He became famous worldwide due to his creations. He was honoured by the Swiss commission of UNESCO. The Rock garden is a masterpiece of stone and recycled material created by the 80 year old Nek Chand. This is one of his works of outsider art or art brut. Outsider art is created by people who have not received formal training in the finer pieces of art.
More information:
This is a prose "Landscape of the Soul" written by one of the famous art experts Nathalie Trouveroy. In this article she differentiates between the eastern and western paintings. A western painting is based on a single viewpoint and is not factual where as the landscape of eastern paintings is spiritual and the viewer can enter it from any angle.
Answer:
Answer:
Explanation:
\Large{\underline{\underline{\it{Given:}}}}
Given:
\sf{\dfrac{tan\:A}{sec\:A-1} -\dfrac{sin\:A}{1+cos\:A} =2\:cot\:A}
secA−1
tanA
−
1+cosA
sinA
=2cotA
\Large{\underline{\underline{\it{To\:Prove:}}}}
ToProve:
LHS = RHS
\Large{\underline{\underline{\it{Solution:}}}}
Solution:
→ Taking the LHS of the equation,
\sf{LHS=\dfrac{tan\:A}{sec\:A-1} -\dfrac{sin\:A}{1+cos\:A} }LHS=
secA−1
tanA
−
1+cosA
sinA
→ Applying identities we get
=\sf{\dfrac{\dfrac{sin\:A}{cos\:A} }{\dfrac{1}{cos\:A}-1 } -\dfrac{sin\:A}{1+cos\:A} }=
cosA
1
−1
cosA
sinA
−
1+cosA
sinA
→ Cross multiplying,
=\sf{\dfrac{\dfrac{sin\:A}{cos\:A} }{\dfrac{1-cos\:A}{cos\:A} } -\dfrac{sin\:A}{1+cos\:A} }=
cosA
1−cosA
cosA
sinA
−
1+cosA
sinA
→ Cancelling cos A on both numerator and denominator
=\sf{\dfrac{sin\:A}{1-cos\:A} -\dfrac{sin\:A}{1+cos\:A}}=
1−cosA
sinA
−
1+cosA
sinA
→ Again cross multiplying we get,
=\sf{\dfrac{sin\:A(1+cos\:A)-sin\:A(1-cos\:A)}{(1+cos\:A)(1-cos\:A)}}=
(1+cosA)(1−cosA)
sinA(1+cosA)−sinA(1−cosA)
→ Taking sin A as common,
\sf{=\dfrac{sin\:A[1+cos\:A-(1-cos\:A)]}{(1^{2}-cos^{2}\:A ) }}=
(1
2
−cos
2
A)
sinA[1+cosA−(1−cosA)]
\sf{=\dfrac{sin\:A[1+cos\:A-1+cos\:A]}{sin^{2}\:A } }=
sin
2
A
sinA[1+cosA−1+cosA]
→ Cancelling sin A on both numerator and denominator
\sf{=\dfrac{2\:cos\:A}{sin\:A} }=
sinA
2cosA
\sf=2\times \dfrac{cos\:A}{sin\:A} }
\sf{=2\:cot\:A}=2cotA
=\sf{RHS}=RHS
→ Hence proved.
\Large{\underline{\underline{\it{Identitites\:used:}}}}
Identititesused:
\sf{tan\:A=\dfrac{sin\:A}{cos\:A} }tanA=
cosA
sinA
\sf{sec\:A=\dfrac{1}{cos\:A} }secA=
cosA
1
\sf{(a+b)\times(a-b)=a^{2}-b^{2} }(a+b)×(a−b)=a
2
−b
2
\sf{(1-cos^{2}\:A)=sin^{2} \:A}(1−cos
2
A)=sin
2
A
\sf{\dfrac{cos\:A}{sin\:A}=cot\:A}
sinA
cosA
=cotA