guys please help me in solving this question sin theta[1+tan theta] + cos theta [1+ cot theta]=sec theta + cosec theta please give answer only About the author Parker
[tex]\large\underline{\bold{Given \:Question – }}[/tex] [tex] \bf \: Prove \: that \: \sf \: sin\theta \:(1 + tan\theta \:) + cos\theta \:(1 + cot\theta \:) = sec\theta \: + cosec\theta \:[/tex] [tex]\begin{gathered}\Large{\bold{{\underline{Formula \: Used – }}}} \end{gathered}[/tex] [tex] \boxed{ \bf{ \: tan\theta \: = \dfrac{sin\theta \:}{cos\theta \:} }}[/tex] [tex] \boxed{ \bf{ \: cot\theta \: = \dfrac{cos\theta \:}{sin\theta \:} }}[/tex] [tex] \boxed{ \bf{ \: \dfrac{1}{cos\theta \:} = sec\theta \:}}[/tex] [tex] \boxed{ \bf{ \: \dfrac{1}{sin\theta \:} = cosec\theta \:}}[/tex] [tex] \boxed{ \bf{ \: {sin}^{2} \theta \: + {cos}^{2} \theta \: = 1}}[/tex] [tex]\large\underline{\sf{Solution-}}[/tex] Consider, [tex]\rm :\longmapsto\:\: sin\theta \:(1 + tan\theta \:) + cos\theta \:(1 + cot\theta \:)[/tex] [tex] \rm \: = \: \: sin\theta \: \bigg(1 + \dfrac{sin\theta \:}{cos\theta \:} \: \bigg) + cos\theta \: \bigg(1 + \dfrac{cos\theta \:}{sin\theta \:} \: \bigg)[/tex] [tex] \rm \: = \: \: sin\theta \: \bigg( \dfrac{cos\theta \: + sin\theta \:}{cos\theta \:} \: \bigg) + cos\theta \: \bigg(\dfrac{sin\theta \: + cos\theta \:}{sin\theta \:} \: \bigg)[/tex] [tex] \rm \: = \: (sin\theta \: + cos\theta \:)\bigg(\dfrac{sin\theta \:}{cos\theta \:} + \dfrac{cos\theta \:}{sin\theta \:} \bigg) [/tex] [tex] \rm \: = \: (sin\theta \: + cos\theta \:)\bigg( \dfrac{ {sin}^{2}\theta \: + {cos}^{2}\theta \: }{sin\theta \:cos\theta \:} \bigg) [/tex] [tex] \rm \: = \: (sin\theta \: + cos\theta \:) \times \dfrac{1}{sin\theta \:cos\theta \:} [/tex] [tex] \rm \: = \: \dfrac{ \cancel{sin\theta \:}}{ \cancel{sin\theta } \: \: cos\theta \:} + \dfrac{ \cancel{cos\theta \:}}{sin\theta \: \cancel{cos\theta \:}} [/tex] [tex] \rm \: = \: \dfrac{1}{cos\theta \:} + \dfrac{1}{sin\theta \:} [/tex] [tex] \rm \: = \: sec\theta \: + cosec\theta \:[/tex] [tex]{\boxed{\boxed{\bf{Hence, Proved}}}}[/tex] Additional Information:- Relationship between sides and T ratios sin θ = Opposite Side/Hypotenuse cos θ = Adjacent Side/Hypotenuse tan θ = Opposite Side/Adjacent Side sec θ = Hypotenuse/Adjacent Side cosec θ = Hypotenuse/Opposite Side cot θ = Adjacent Side/Opposite Side Reciprocal Identities cosec θ = 1/sin θ sec θ = 1/cos θ cot θ = 1/tan θ sin θ = 1/cosec θ cos θ = 1/sec θ tan θ = 1/cot θ Co-function Identities sin (90°−x) = cos x cos (90°−x) = sin x tan (90°−x) = cot x cot (90°−x) = tan x sec (90°−x) = cosec x cosec (90°−x) = sec x Fundamental Trigonometric Identities sin²θ + cos²θ = 1 sec²θ – tan²θ = 1 cosec²θ – cot²θ = 1 Reply
[tex]\large\underline{\bold{Given \:Question – }}[/tex]
[tex] \bf \: Prove \: that \: \sf \: sin\theta \:(1 + tan\theta \:) + cos\theta \:(1 + cot\theta \:) = sec\theta \: + cosec\theta \:[/tex]
[tex]\begin{gathered}\Large{\bold{{\underline{Formula \: Used – }}}} \end{gathered}[/tex]
[tex] \boxed{ \bf{ \: tan\theta \: = \dfrac{sin\theta \:}{cos\theta \:} }}[/tex]
[tex] \boxed{ \bf{ \: cot\theta \: = \dfrac{cos\theta \:}{sin\theta \:} }}[/tex]
[tex] \boxed{ \bf{ \: \dfrac{1}{cos\theta \:} = sec\theta \:}}[/tex]
[tex] \boxed{ \bf{ \: \dfrac{1}{sin\theta \:} = cosec\theta \:}}[/tex]
[tex] \boxed{ \bf{ \: {sin}^{2} \theta \: + {cos}^{2} \theta \: = 1}}[/tex]
[tex]\large\underline{\sf{Solution-}}[/tex]
Consider,
[tex]\rm :\longmapsto\:\: sin\theta \:(1 + tan\theta \:) + cos\theta \:(1 + cot\theta \:)[/tex]
[tex] \rm \: = \: \: sin\theta \: \bigg(1 + \dfrac{sin\theta \:}{cos\theta \:} \: \bigg) + cos\theta \: \bigg(1 + \dfrac{cos\theta \:}{sin\theta \:} \: \bigg)[/tex]
[tex] \rm \: = \: \: sin\theta \: \bigg( \dfrac{cos\theta \: + sin\theta \:}{cos\theta \:} \: \bigg) + cos\theta \: \bigg(\dfrac{sin\theta \: + cos\theta \:}{sin\theta \:} \: \bigg)[/tex]
[tex] \rm \: = \: (sin\theta \: + cos\theta \:)\bigg(\dfrac{sin\theta \:}{cos\theta \:} + \dfrac{cos\theta \:}{sin\theta \:} \bigg) [/tex]
[tex] \rm \: = \: (sin\theta \: + cos\theta \:)\bigg( \dfrac{ {sin}^{2}\theta \: + {cos}^{2}\theta \: }{sin\theta \:cos\theta \:} \bigg) [/tex]
[tex] \rm \: = \: (sin\theta \: + cos\theta \:) \times \dfrac{1}{sin\theta \:cos\theta \:} [/tex]
[tex] \rm \: = \: \dfrac{ \cancel{sin\theta \:}}{ \cancel{sin\theta } \: \: cos\theta \:} + \dfrac{ \cancel{cos\theta \:}}{sin\theta \: \cancel{cos\theta \:}} [/tex]
[tex] \rm \: = \: \dfrac{1}{cos\theta \:} + \dfrac{1}{sin\theta \:} [/tex]
[tex] \rm \: = \: sec\theta \: + cosec\theta \:[/tex]
[tex]{\boxed{\boxed{\bf{Hence, Proved}}}}[/tex]
Additional Information:-
Relationship between sides and T ratios
sin θ = Opposite Side/Hypotenuse
cos θ = Adjacent Side/Hypotenuse
tan θ = Opposite Side/Adjacent Side
sec θ = Hypotenuse/Adjacent Side
cosec θ = Hypotenuse/Opposite Side
cot θ = Adjacent Side/Opposite Side
Reciprocal Identities
cosec θ = 1/sin θ
sec θ = 1/cos θ
cot θ = 1/tan θ
sin θ = 1/cosec θ
cos θ = 1/sec θ
tan θ = 1/cot θ
Co-function Identities
sin (90°−x) = cos x
cos (90°−x) = sin x
tan (90°−x) = cot x
cot (90°−x) = tan x
sec (90°−x) = cosec x
cosec (90°−x) = sec x
Fundamental Trigonometric Identities
sin²θ + cos²θ = 1
sec²θ – tan²θ = 1
cosec²θ – cot²θ = 1
Answer:
kya tum meri gf banogi
Step-by-step explanation:
i love you