1 ) Which one is a cardinal number
A) 10
BV
O O O O
C) VII
D) XI
Answers
Answer:
To Proof:
\sf\dfrac{1 + cos \theta - {sin}^{2} \theta}{sin \theta(1 + cos \theta)} sinθ(1+cosθ)
sinθ(1+cosθ)
1+cosθ−sin
2
θ
sinθ(1+cosθ)
1+cosθ−sin 2 θ
\bold{\underline{Proof:}}
Proof:
LHS:
\begin{gathered}\begin{gathered}\begin{gathered}\sf\red\implies \dfrac{1 + cos \theta - {sin}^{2} \theta}{sin \theta(1 + cos \theta)} \\ \\ \sf {sin}^{2} \theta = 1 - {cos}^{2} \theta : \\ \sf \pink\implies \dfrac{1 + cos \theta - (1 - {cos}^{2} \theta )}{sin \theta(1 + cos \theta)} \\ \\ \sf\green(1 - cos ^{2} \theta ) = ( {1}^{2} - cos ^{2} \theta ) = (1 - cos \theta )(1 + cos \theta ) : \\ \sf\purple \implies \dfrac{1 + cos \theta - (1 - cos \theta )(1 + cos \theta )}{sin \theta(1 + cos \theta)} \\ \\ \sf \implies \dfrac{ \cancel{1 + cos \theta}(1 - (1 - cos \theta ))}{sin \theta( \cancel{1 + cos \theta})} \\ \\ \sf\orange \implies \dfrac{1 - (1 - cos \theta )}{sin \theta} \\ \\ \sf\red \implies \dfrac{1 - 1 + cos \theta )}{sin \theta} \\ \\ \sf\blue \implies\dfrac{cos \theta }{sin \theta} \\ \\ \sf\pink \implies cot \theta\end{gathered}\end{gathered}\end{gathered}
⟹
sinθ(1+cosθ)
1+cosθ−sin
2
θ
sin
2
θ=1−cos
2
θ:
⟹
sinθ(1+cosθ)
1+cosθ−(1−cos
2
θ)
(1−cos
2
θ)=(1
2
−cos
2
θ)=(1−cosθ)(1+cosθ):
⟹
sinθ(1+cosθ)
1+cosθ−(1−cosθ)(1+cosθ)
⟹
sinθ(
1+cosθ
)
1+cosθ
(1−(1−cosθ))
⟹
sinθ
1−(1−cosθ)
⟹
sinθ
1−1+cosθ)
⟹
sinθ
cosθ
⟹cotθ
⟹ sinθ(1+cosθ) 1+cosθ−sin 2 θ sin 2θ=1−cos 2θ:⟹sinθ(1+cosθ)1+cosθ−sin2θsin2θ=1−cos2θ:
⟹ sinθ(1+cosθ) 1+cosθ−(1−cos2θ) (1−cos 2θ)=(1 2−cos 2 θ)=(1−cosθ)(1+cosθ):⟹sinθ(1+cosθ)1+cosθ−(1−cos2θ)(1−cos2θ)=(12−cos2θ)=(1−cosθ)(1+cosθ):
⟹ sinθ(1+cosθ) 1+cosθ−(1−cosθ)(1+cosθ)⟹sinθ(1+cosθ)1+cosθ−(1−cosθ)(1+cosθ)
⟹ sinθ( 1+cos0)1+cosθ (1−(1−cosθ))⟹sinθ(1+cos0)1+cosθ(1−(1−cosθ))
⟹ sinθ1−(1−cosθ)⟹sinθ1−(1−cosθ)
⟹ sinθ 1−1+cosθ)⟹sinθ1−1+cosθ)
⟹ sinθ cosθ⟹sinθcosθ
⟹cotθ⟹cotθ
RHS:
\sf\green\implies cot \theta⟹cotθ⟹cotθ⟹cotθ
\therefore∴∴∴
\bold\red{LHS = RHS}LHS=RHS
\sf\pink{Hence \: Proved}HenceProved
Answer:
10 is a cardinal number
pls mark my answer brainliest