if cosec theta + cot theta =m show that m square -1÷ m square + 1 =0
Answers
Answer:
Step-by-step explanation:
Answer:-
\huge{\rm{\pink{\underline{Given:-}}}}
Given:−
\sf{Cosec \theta + Cot \theta = m}Cosecθ+Cotθ=m
\huge{\rm{\pink{\underline{To \ prove:-}}}}
To prove:−
\sf{(m^2+1)Cos \theta = m^2 -1}(m
2
+1)Cosθ=m
2
−1
Concept:-
Trigonometry and its applications.
Let's Do!
\sf{(m^2+1)Cos \theta = m^2 -1}(m
2
+1)Cosθ=m
2
−1
\sf{Cos \theta = \dfrac{m^2 -1}{(m^2+1)}}Cosθ=
(m
2
+1)
m
2
−1
(1)
\sf{Cosec \theta + Cot \theta = m}Cosecθ+Cotθ=m
\sf{\dfrac{1}{sin \theta} + \dfrac{cos \theta }{sin \theta} = m}
sinθ
1
+
sinθ
cosθ
=m
\sf{\dfrac{1 + cos \theta}{sin \theta} = m}
sinθ
1+cosθ
=m (2)
Placing value of (2) in (1),
Numerator:-
\sf{(\dfrac{1+ cos \theta}{sin \theta})^2 - 1 }(
sinθ
1+cosθ
)
2
−1
\sf{\dfrac{(1+ cos \theta)^2}{sin^2 \theta} - 1 }
sin
2
θ
(1+cosθ)
2
−1
\sf{\dfrac{(1 + cos^2 \theta+2cos \theta)}{sin^2 \theta} - 1 }
sin
2
θ
(1+cos
2
θ+2cosθ)
−1
\sf{\dfrac{1 + cos^2 \theta+2cos \theta - sin^2 \theta}{sin^2 \theta}}
sin
2
θ
1+cos
2
θ+2cosθ−sin
2
θ
\sf{\dfrac{cos^2 \theta + cos^2 \theta+2cos \theta }{sin^2 \theta}}
sin
2
θ
cos
2
θ+cos
2
θ+2cosθ
\sf{\dfrac{2cos^2 \theta +2cos \theta }{sin^2 \theta}}
sin
2
θ
2cos
2
θ+2cosθ
Denominator:-
\sf{(\dfrac{1+ cos \theta}{sin \theta})^2 + 1 }(
sinθ
1+cosθ
)
2
+1
\sf{\dfrac{(1+ cos \theta)^2}{sin^2 \theta} + 1 }
sin
2
θ
(1+cosθ)
2
+1
\sf{\dfrac{(1 + cos^2 \theta+2cos \theta)}{sin^2 \theta} + 1 }
sin
2
θ
(1+cos
2
θ+2cosθ)
+1
\sf{\dfrac{1 + cos^2 \theta+2cos \theta + sin^2 \theta}{sin^2 \theta}}
sin
2
θ
1+cos
2
θ+2cosθ+sin
2
θ
\sf{\dfrac{1 +1+2cos \theta}{sin^2 \theta}}
sin
2
θ
1+1+2cosθ
\sf{\dfrac{2+2cos \theta}{sin^2 \theta}}
sin
2
θ
2+2cosθ
Now, combining both:-
\sf{\dfrac{\dfrac{2cos^2 \theta +2cos \theta }{sin^2 \theta}}{\dfrac{2+2cos \theta}{sin^2 \theta}}}
sin
2
θ
2+2cosθ
sin
2
θ
2cos
2
θ+2cosθ
Now, sin² theta gets cancelled.
\sf{\dfrac{2 cos^2 \theta + 2 cos \theta }{2+2 cos \theta}}
2+2cosθ
2cos
2
θ+2cosθ
\sf{\dfrac{2 cos \theta( cos \theta +1)}{2(1+ cos \theta)}}
2(1+cosθ)
2cosθ(cosθ+1)
2 gets cancelled, 1 + cos theta cancelled, so left is cos theta.
Hence Proved