7. Did you like the nonsense poems? How are these po
ems different
from the other poems you have learnt?
Answers
Answer:
Prove that:
\displaystyle\sf\:\dfrac{\csc\:\theta}{\csc\:\theta\:-\:1}\:+\:\dfrac{\csc\:\theta}{\csc\:\theta\:+\:1}\:=\:\dfrac{2}{\cos^2\:\theta}
cscθ−1
cscθ
+
cscθ+1
cscθ
=
cos
2
θ
2
Answer:
\displaystyle\boxed{\red{\sf\:\dfrac{\csc\:\theta}{\csc\:\theta\:-\:1}\:+\:\dfrac{\csc\:\theta}{\csc\:\theta\:+\:1}\:=\:\dfrac{2}{\cos^2\:\theta}}}
cscθ−1
cscθ
+
cscθ+1
cscθ
=
cos
2
θ
2
Step-by-step-explanation:
We have to prove the given trigonometric equation.
Considering LHS of the given equation,
\displaystyle{\implies\sf\:LHS\:=\:\dfrac{\csc\:\theta}{\csc\:\theta\:-\:1}\:+\:\dfrac{\csc\:\theta}{\csc\:\theta\:+\:1}}⟹LHS=
cscθ−1
cscθ
+
cscθ+1
cscθ
\displaystyle{\implies\sf\:LHS\:=\:\dfrac{\csc\:\theta\:\times\:(\:\csc\:\theta\:+\:1\:)\:+\:\csc\:\theta\:\times\:(\:\csc\:\theta\:-\:1\:)}{\:(\:\csc\:\theta\:-\:1\:)\:\times\:(\:\csc\:\theta\:+\:1\:)}}⟹LHS=
(cscθ−1)×(cscθ+1)
cscθ×(cscθ+1)+cscθ×(cscθ−1)
\displaystyle{\implies\sf\:LHS\:=\:\dfrac{\csc^2\:\theta\:+\:1\:+\:\csc^2\:\theta\:-\:1}{\csc^2\:\theta\:-\:1^2}\:\:\:-\:-\:[\:(\:a\:+\:b\:)\:(\:a\:-\:b\:)\:=\:a^2\:-\:b^2\:]}⟹LHS=
csc
2
θ−1
2
csc
2
θ+1+csc
2
θ−1
−−[(a+b)(a−b)=a
2
−b
2
]
\displaystyle{\implies\sf\:LHS\:=\:\dfrac{\csc^2\:\theta\:+\:\csc^2\:\theta}{\csc^2\:\theta\:-\:1}}⟹LHS=
csc
2
θ−1
csc
2
θ+csc
2
θ
\displaystyle{\implies\sf\:LHS\:=\:\dfrac{2\:\csc^2\:\theta}{\cot^2\:\theta}\:\:\:-\:-\:[\:\because\:1\:+\:\cot^2\:\theta\:=\:\csc^2\:\theta\:]}⟹LHS=
cot
2
θ
2csc
2
θ
−−[∵1+cot
2
θ=csc
2
θ]
\displaystyle{\implies\sf\:LHS\:=\:\dfrac{2\:\times\:\left(\:\dfrac{1}{\sin\:\theta}\:\right)^2}{\cot^2\:\theta}\:\:\:-\:-\:-\:[\:\because\:\csc\:\theta\:=\:\dfrac{1}{\sin\:\theta}\:]}⟹LHS=
cot
2
θ
2×(
sinθ
1
)
2
−−−[∵cscθ=
sinθ
1
]
\displaystyle{\implies\sf\:LHS\:=\:\dfrac{2\:\times\:\left(\:\dfrac{1}{\sin\:\theta}\:\right)^2}{\left(\:\dfrac{\cos\:\theta}{\sin\:\theta}\:\right)^2}\:\:\:-\:-\:-\:[\:\because\:\cot\:\theta\:=\:\dfrac{\cos\:\theta}{\sin\:\theta}\:]}⟹LHS=
(
sinθ
cosθ
)
2
2×(
sinθ
1
)
2
−−−[∵cotθ=
sinθ
cosθ
]
\displaystyle{\implies\sf\:LHS\:=\:\dfrac{2\:\times\:\left(\:\dfrac{1}{\sin\:\theta}\:\right)\:\times\:\left(\:\dfrac{1}{\sin\:\theta}\:\right)}{\left(\:\dfrac{\cos\:\theta}{\sin\:\theta}\:\right)\:\times\:\left(\:\dfrac{\cos\:\theta}{\sin\:\theta}\:\right)}}⟹LHS=
(
sinθ
cosθ
)×(
sinθ
cosθ
)
2×(
sinθ
1
)×(
sinθ
1
)
\displaystyle{\implies\sf\:LHS\:=\:\dfrac{2\:\times\:\cancel{\sin\:\theta}\:\times\:\cancel{\sin\:\theta}}{\cancel{\sin\:\theta}\:\times\:\cancel{\sin\:\theta}\:\times\:\cos\:\theta\:\times\:\cos\:\theta}}⟹LHS=
sinθ
×
sinθ
×cosθ×cosθ
2×
sinθ
×
sinθ
\displaystyle{\implies\sf\:LHS\:=\:\dfrac{2}{\cos^2\:\theta}}⟹LHS=
cos
2
θ
2
\displaystyle{\implies\sf\:RHS\:=\:\dfrac{2}{\cos^2\:\theta}}⟹RHS=
cos
2
θ
2
\displaystyle{\therefore\boxed{\red{\sf\:LHS\:=\:RHS}}}∴
LHS=RHS
Hence proved!