PROVE
(secø - cosø)(cotø+tanø) = tanø secø
Answers
Step-by-step explanation:
Used formulae:-
- secA=1/CosA
- Tan A=SinA/CosA
- CotA=CosA/SinA
- sin²A+Cos²A=1
- sec²A+tan²A=1
Step-by-step explanation:
꧁
answer
꧂
resistors :- 6 ohm and 12 ohm
voltage :- 6 V
\bold { \underline{ \underline \orange{answer}}} \orange→
answer
→
2:9
\bold { \underline{ \underline \orange{solution}}} \orange→
solution
→
case (i)
when resistors are connected in series ;
Rs = R1 + R2
→ Rs = 6 + 12
→ Rs = 18 ohm
so, effective resistance in series combination is 18 ohm
we know that..
\begin{gathered} \bold{ \color{blue} \: h = \frac{ {v}^{2} t}{r}} \\ \end{gathered}
h=
r
v
2
t
\begin{gathered} \bold{ \implies \: h1 = \frac{ {6}^{2}t }{18} } \\ \\ \bold{ \implies h1 = \frac{36t}{18} } \\ \\ \bold{\implies \: h1 = 2t} \: \: \: \end{gathered}
⟹h1=
18
6
2
t
⟹h1=
18
36t
⟹h1=2t
so heat produced in series combination is 2t joule
case (ii)
when resistors are connected in parallel
we know that;
\begin{gathered} \bold{ \frac{1}{Rp} = \frac{1}{r1} + \frac{1}{r2} } \\ \end{gathered}
Rp
1
=
r1
1
+
r2
1
\begin{gathered} \bold { \tt{→ \: \frac{1}{rp} = \frac{1}{6} + \frac{1}{12} }} \: \: \: \: \: \: \\ \\ \bold{ \tt \implies \: \frac{1}{Rp} = \frac{3}{12} } \: \: \: \: \: \: \: \: \: \\ \\ \bold{ \implies \: Rp = \frac{12}{3} } \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bold{\implies \: Rp = 4 \: ohm } \: \: \: \: \end{gathered}
→
rp
1
=
6
1
+
12
1
⟹
Rp
1
=
12
3
⟹Rp=
3
12
⟹Rp=4ohm
so, effective resistance in parallel combination is 4 ohm
and
\begin{gathered} \bold{ \color{blue} \: h = \frac{ {v}^{2} t}{r}} \\\end{gathered}
h=
r
v
2
t
\begin{gathered} \bold{ \implies \: h2 = \frac{ {6}^{2}t }{4} } \\ \\ \bold{ \implies \: h2 = \frac{36t}{4} } \\ \\ \bold{ \implies \: h2 = 9t} \: \: \: \end{gathered}
⟹h2=
4
6
2
t
⟹h2=
4
36t
⟹h2=9t
so heat produced in parallel combination is 9t joule.
\begin{gathered} \bold{ \color{pink}ratio \: of \: h1 \: and \: h2 \: } \\ \\ \bold \blue{ \frac{h1}{h2} = \frac{2t}{9t} } \\ \\ \bold{so \: \: \: h1 : h2 =2 : 9 }\end{gathered}
ratioofh1andh2
h2
h1
=
9t
2t
soh1:h2=2:9
hence, ratio of heat produced in the series combination to that of the parallel combination of resistors is 2:9
id of moderators and brainly stars;
Shadowsabers03
Itzfaded Elena
Legendrenaline
no4
Sk2
Brainly Ronaldo
Kalpeshprabhakar
Nikki57
Arya1308
Tanu81
kvnmurty
rsagnik437
Rythm14
Rose08
collestcat 015
Themoonlightphoenix
Ar17
praxi13
HarishAS
Praneethworldtopper꧁
answer
꧂
resistors :- 6 ohm and 12 ohm
voltage :- 6 V
\bold { \underline{ \underline \orange{answer}}} \orange→
answer
→
2:9
\bold { \underline{ \underline \orange{solution}}} \orange→
solution
→
case (i)
when resistors are connected in series ;
Rs = R1 + R2
→ Rs = 6 + 12
→ Rs = 18 ohm
so, effective resistance in series combination is 18 ohm
we know that..
\begin{gathered} \bold{ \color{blue} \: h = \frac{ {v}^{2} t}{r}} \\ \end{gathered}
h=
r
v
2
t
\begin{gathered} \bold{ \implies \: h1 = \frac{ {6}^{2}t }{18} } \\ \\ \bold{ \implies h1 = \frac{36t}{18} } \\ \\ \bold{\implies \: h1 = 2t} \: \: \: \end{gathered}
⟹h1=
18
6
2
t
⟹h1=
18
36t
⟹h1=2t
so heat produced in series combination is 2t joule
case (ii)
when resistors are connected in parallel
we know that;
\begin{gathered} \bold{ \frac{1}{Rp} = \frac{1}{r1} + \frac{1}{r2} } \\ \end{gathered}
Rp
1
=
r1
1
+
r2
1
\begin{gathered} \bold { \tt{→ \: \frac{1}{rp} = \frac{1}{6} + \frac{1}{12} }} \: \: \: \: \: \: \\ \\ \bold{ \tt \implies \: \frac{1}{Rp} = \frac{3}{12} } \: \: \: \: \: \: \: \: \: \\ \\ \bold{ \implies \: Rp = \frac{12}{3} } \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bold{\implies \: Rp = 4 \: ohm } \: \: \: \: \end{gathered}
→
rp
1
=
6
1
+
12
1
⟹
Rp
1
=
12
3
⟹Rp=
3
12
⟹Rp=4ohm
so, effective resistance in parallel combination is 4 ohm
and
\begin{gathered} \bold{ \color{blue} \: h = \frac{ {v}^{2} t}{r}} \\\end{gathered}
h=
r
v
2
t
\begin{gathered} \bold{ \implies \: h2 = \frac{ {6}^{2}t }{4} } \\ \\ \bold{ \implies \: h2 = \frac{36t}{4} } \\ \\ \bold{ \implies \: h2 = 9t} \: \: \: \end{gathered}
⟹h2=
4
6
2
t
⟹h2=
4
36t
⟹h2=9t
so heat produced in parallel combination is 9t joule.
\begin{gathered} \bold{ \color{pink}ratio \: of \: h1 \: and \: h2 \: } \\ \\ \bold \blue{ \frac{h1}{h2} = \frac{2t}{9t} } \\ \\ \bold{so \: \: \: h1 : h2 =2 : 9 }\end{gathered}
ratioofh1andh2
h2
h1
=
9t
2t
soh1:h2=2:9
hence, ratio of heat produced in the series combination to that of the parallel combination of resistors is 2:9
id of moderators and brainly stars;
Shadowsabers03
Itzfaded
\begin{gathered} \bold{ \color{blue} \: h = \frac{ {v}^{2} t}{r}} \\ \end{gathered}
h=
r
v
2
t
\begin{gathered} \bold{ \implies \: h1 = \frac{ {6}^{2}t }{18} } \\ \\ \bold{ \implies h1 = \frac{36t}{18} } \\ \\ \bold{\implies \: h1 = 2t} \: \: \: \end{gathered}
⟹h1=
18
6
2
t
⟹h1=
18
36t
⟹h1=2t
so heat produced in series combination is 2t joule
case
Ar17
praxi13
HarishAS
Praneethworldtopper