Hye gays anyone need help.
Answers
Answer:
डोकरी ढुमका टैक्स
Explanation:
Step-by-step explanation:
★Given -
\bf \rm{ \dfrac{sec \theta \: + tan \theta}{sec \theta \: - tan \theta} }
secθ−tanθ
secθ+tanθ
★To prove -
\bf\rm{ {\huge(} \dfrac{1 + sin \theta}{cos \theta} {\huge)}^{2} }(
cosθ
1+sinθ
)
2
★Solution -
L.H.S
\longmapsto \bf \rm{ \dfrac{sec \theta \: + tan \theta}{sec \theta \: - tan \theta} }⟼
secθ−tanθ
secθ+tanθ
\longmapsto \bf \rm{ \dfrac{ \dfrac{1}{cos \theta} \: + \dfrac{sin \theta}{cos \theta} }{ \dfrac{1}{cos \theta} \: - \dfrac{sin \theta}{cos \theta} } }⟼
cosθ
1
−
cosθ
sinθ
cosθ
1
+
cosθ
sinθ
\bf \longmapsto \rm{ \dfrac{ \dfrac{1 + sin \theta}{cos \theta} }{ \dfrac{1 - sin \theta}{cos \theta} } }⟼
cosθ
1−sinθ
cosθ
1+sinθ
\longmapsto \bf \rm{ \dfrac{1 + sin \theta}{ \cancel{cos \theta}} \times \dfrac{ \cancel{cos \theta}}{ 1 - sin \theta} }⟼
cosθ
1+sinθ
×
1−sinθ
cosθ
\longmapsto \bf \rm{ \dfrac{1 + sin \theta}{1 - sin \theta}}⟼
1−sinθ
1+sinθ
\longmapsto \bf \rm{ \dfrac{1 + sin \theta}{1 - sin \theta} \times \dfrac{1 + sin \theta}{1 + sin \theta} }⟼
1−sinθ
1+sinθ
×
1+sinθ
1+sinθ
\longmapsto \bf \rm{ \dfrac{(1 + sin \theta)^{2} }{ {1}^{2} - sin^{2} \theta}}⟼
1
2
−sin
2
θ
(1+sinθ)
2
\longmapsto \bf \rm{ \dfrac{(1 + sin \theta)^{2} }{ {1} - sin^{2} \theta}}⟼
1−sin
2
θ
(1+sinθ)
2
∵ sin²∅ + cos ²∅ = 1
=> cos²∅ = 1 - sin²∅
\longmapsto \bf \rm{ \dfrac{(1 + sin \theta)^{2} }{ {cos}^{2} \theta}}⟼
cos
2
θ
(1+sinθ)
2
\longmapsto \bf \rm{ \dfrac{(1 + sin \theta)^{2} }{ ( {cos} \theta) ^{2} }}⟼
(cosθ)
2
(1+sinθ)
2
\longmapsto \bf \green{\rm{ { {{\huge(} \dfrac{1 + sin \theta}{cos \theta} {\huge)}^{2} }}}= R.H.S}⟼(
cosθ
1+sinθ
)
2
=R.H.S
★More to know -
\rm{sin \theta = \dfrac{1}{cosec \theta} }sinθ=
cosecθ
1
\rm{cos \theta = \dfrac{1}{sec \theta} }cosθ=
secθ
1
\rm{tan \theta = \dfrac{1}{cot \theta} }tanθ=
cotθ
1
\rm{cosec \theta = \dfrac{1}{sin \theta} }cosecθ=
sinθ
1
\rm{sec \theta = \dfrac{1 }{cos \theta} }secθ=
cosθ
1
\rm{cot \theta = \dfrac{1 }{tan \theta} }cotθ=
tanθ
1
Yassss !!!
Colloids are classified into two types in the cases where dispersion medium is water. They are:
(i) Hydrophilic or Lyophilic colloids : Those substances, which when mixed with the dispersion medium, form directly the colloidal solution and are termed as hydrophilic colloids. They are reversible solutions, quite stable and cannot be easily precipitated.
Example : gum, gelatine, starch, rubber etc.
(ii) Hydrophobic or Lyophobic colloids : Those substances which do not form colloidal solution when simply mixed with the dispersion medium, are called hydrophobic colloids. They are irreversible solutions, unstable and can be easily precipitated. Example : Metals and their sulphides