f v = log(tan x + tan y + tan z) , find the value of : z v z dv dy y x v x ∂ ∂ + + ∂ ∂ sin 2 sin 2 sin 2 .
Answers
Step-by-step explanation:
༒ To Prove :-
\begin{gathered} \Large\bf( {1 - tanA)}^{2} (1 - {cotA)}^{2} \\\Large= \bf(secA - cosecA) {}^{2}\end{gathered}
(1−tanA)
2
(1−cotA)
2
=(secA−cosecA)
2
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༒ Formulae Used :-
\begin{gathered} \bf \maltese \: \: \: 1 + {tan}^{2} A = sec {}^{2} A \\ \\ \bf \maltese \: \: \: 1 + {cot} {}^{2} A = {cosec}^{2} A \\ \\ \bf \maltese \: \: \: sin^2\theta+cos^2\theta=1 \\ \\ \bf \maltese \: \: \: \frac{1}{sin\theta}=cosec\theta \\ \\ \bf \maltese \: \: \: \frac{1}{cos\theta}=sec\theta\end{gathered}
✠1+tan
2
A=sec
2
A
✠1+cot
2
A=cosec
2
A
✠sin
2
θ+cos
2
θ=1
✠
sinθ
1
=cosecθ
✠
cosθ
1
=secθ
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༒ Proof :-
\begin{gathered} \sf {(1 - tanA) }^{2} + {(1 - cot A)}^{2} \\ \\ \sf = (1 + {tan}^{2}A - 2tanA) \\ \sf+ ( 1 + {cot}^{2} A - 2cotA) \\ \\ \sf = {sec}^{2} A - 2tanA + {cosec}^{2} A - 2cotA \\ \{ \bf\because1 + {tan}^{2} A = sec {}^{2} A \\ \bf and1 + {cot} {}^{2} A = {cosec}^{2} A \} \\ \\ = \sf {sec}^{2} A + {cosec}^{2} A - 2( tanA + cotA) \\ \\ = \sf {sec}^{2} A + {cosec}^{2} A - 2( \frac{sinA }{cosA }+ \frac{cosA}{sinA} ) \\ \\ \sf = {sec}^{2} A + {cosec}^{2} A - 2( \frac{ {sin}^{2} A + {cos}^{2}A)}{cosA \: sin A } \\ \\ = \sf {sec}^{2} + {cosec}^{2} A - 2( \frac{1}{sinA \: cos A } )\\\{\bf\because sin^2\theta+cos^2\theta=1\} \\ \\ \sf = {sec}^{2} A + {cosec}^{2} A - 2secA \: cosecA \\\\ \{\bf\because\frac{1}{sin\theta}=cosec\theta\\ \bf and\:\:\frac{1}{cos\theta}=sec\theta\} \\ \\ \sf = (secA - cosecA) {}^{2} \end{gathered}
(1−tanA)
2
+(1−cotA)
2
=(1+tan
2
A−2tanA)
+(1+cot
2
A−2cotA)
=sec
2
A−2tanA+cosec
2
A−2cotA
{∵1+tan
2
A=sec
2
A
and1+cot
2
A=cosec
2
A}
=sec
2
A+cosec
2
A−2(tanA+cotA)
=sec
2
A+cosec
2
A−2(
cosA
sinA
+
sinA
cosA
)
=sec
2
A+cosec
2
A−2(
cosAsinA
sin
2
A+cos
2
A)
=sec
2
+cosec
2
A−2(
sinAcosA
1
)
{∵sin
2
θ+cos
2
θ=1}
=sec
2
A+cosec
2
A−2secAcosecA
{∵
sinθ
1
=cosecθ
and
cosθ
1
=secθ}
=(secA−cosecA)
2
\begin{gathered} \Large \red{\mathfrak{ \text{W}hich \: \: is \: \: the \: \: required} }\\ \huge \red{\mathfrak{ \text{ A}nswer.}}\end{gathered}
Which is the required
Answer.
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