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$increment A B C comma$ if $sin to the power of 2 end exponent invisible function application fraction numerator A over denominator 2 end fraction comma sin to the power of 2 end exponent invisible function application fraction numerator B over denominator 2 end fraction$ and $sin to the power of 2 end exponent invisible function application fraction numerator C over denominator 2 end fraction$ are in H.P., then $blank a comma blank b blank$ and $blank c blank$ will be in

A.P.
G.P.
H.P.
None of these

The correct answer is: H.P.

$fraction numerator 1 over denominator sin to the power of 2 end exponent invisible function application fraction numerator A over denominator 2 end fraction end fraction comma fraction numerator 1 over denominator sin to the power of 2 end exponent invisible function application fraction numerator B over denominator 2 end fraction end fraction comma fraction numerator 1 over denominator sin to the power of 2 end exponent invisible function application fraction numerator C over denominator 2 end fraction end fraction blank a r e blank i n blank A. P.$
$rightwards double arrow fraction numerator 1 over denominator sin to the power of 2 end exponent invisible function application fraction numerator C over denominator 2 end fraction end fraction minus fraction numerator 1 over denominator sin to the power of 2 end exponent invisible function application fraction numerator B over denominator 2 end fraction end fraction equals fraction numerator 1 over denominator sin to the power of 2 end exponent invisible function application fraction numerator B over denominator 2 end fraction end fraction minus fraction numerator 1 over denominator sin to the power of 2 end exponent invisible function application fraction numerator A over denominator 2 end fraction end fraction$
$rightwards double arrow fraction numerator a b over denominator open parentheses s minus a close parentheses open parentheses s minus b close parentheses end fraction minus fraction numerator a c over denominator open parentheses s minus a close parentheses open parentheses s minus c close parentheses end fraction equals fraction numerator a c over denominator open parentheses s minus a close parentheses open parentheses s minus c close parentheses end fraction minus fraction numerator b c over denominator open parentheses s minus b close parentheses open parentheses s minus c close parentheses end fraction$
$rightwards double arrow open parentheses fraction numerator a over denominator s minus a end fraction close parentheses open parentheses fraction numerator b open parentheses s minus c close parentheses minus c open parentheses s minus b close parentheses over denominator open parentheses s minus b close parentheses open parentheses s minus c close parentheses end fraction close parentheses equals open parentheses fraction numerator c over denominator s minus c end fraction close parentheses open parentheses fraction numerator a open parentheses s minus b close parentheses minus b open parentheses s minus a close parentheses over denominator open parentheses s minus a close parentheses open parentheses s minus b close parentheses end fraction close parentheses$
$rightwards double arrow a b minus a c equals a c minus b c$
$rightwards double arrow a b plus b c equals 2 a c$
$rightwards double arrow fraction numerator 1 over denominator c end fraction plus fraction numerator 1 over denominator a end fraction equals fraction numerator 2 over denominator b end fraction comma blank i. e. comma blank a comma blank b comma blank c blank$ are in H.P.

Related Questions to study

Member of solutions of $tan invisible function application open parentheses fraction numerator pi over denominator 2 end fraction sin invisible function application theta close parentheses equals cot invisible function application open parentheses fraction numerator pi over denominator 2 end fraction cos invisible function application theta close parentheses comma blank theta element of open square brackets 0 comma blank 6 pi close square brackets comma blank$ is

Member of solutions of $tan invisible function application open parentheses fraction numerator pi over denominator 2 end fraction sin invisible function application theta close parentheses equals cot invisible function application open parentheses fraction numerator pi over denominator 2 end fraction cos invisible function application theta close parentheses comma blank theta element of open square brackets 0 comma blank 6 pi close square brackets comma blank$ is

$i s blank e q u a l blank t o$

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$sin to the power of 2 end exponent invisible function application theta equals fraction numerator x to the power of 2 end exponent plus y to the power of 2 end exponent plus 1 over denominator 2 x end fraction comma blank$ then $x blank$ must be

$sin to the power of 2 end exponent invisible function application theta equals fraction numerator x to the power of 2 end exponent plus y to the power of 2 end exponent plus 1 over denominator 2 x end fraction comma blank$ then $x blank$ must be

parallel

e equation $cos to the power of 8 end exponent invisible function application x plus b cos to the power of 4 end exponent invisible function application x plus 1 equals 0$ will have a solution if $b$ belongs to

e equation $cos to the power of 8 end exponent invisible function application x plus b cos to the power of 4 end exponent invisible function application x plus 1 equals 0$ will have a solution if $b$ belongs to

e minimum vertical distance between the graphs of $y equals 2 plus sin invisible function application x a n d blank y equals cos invisible function application x blank i s$

e minimum vertical distance between the graphs of $y equals 2 plus sin invisible function application x a n d blank y equals cos invisible function application x blank i s$

e number of solutions of $2 sin to the power of 2 end exponent invisible function application x plus sin to the power of 2 end exponent invisible function application 2 x equals 2 comma blank x element of open square brackets 0 comma blank 2 pi close square brackets$ is

e number of solutions of $2 sin to the power of 2 end exponent invisible function application x plus sin to the power of 2 end exponent invisible function application 2 x equals 2 comma blank x element of open square brackets 0 comma blank 2 pi close square brackets$ is

parallel

e general solution of $tan invisible function application theta plus tan invisible function application 2 theta plus tan invisible function application 3 theta equals 0$ is

e general solution of $tan invisible function application theta plus tan invisible function application 2 theta plus tan invisible function application 3 theta equals 0$ is

$sin invisible function application open parentheses alpha plus beta close parentheses equals 1 comma sin invisible function application open parentheses alpha minus beta close parentheses equals fraction numerator 1 over denominator 2 end fraction comma blank$ then $tan invisible function application open parentheses alpha plus 2 beta close parentheses tan invisible function application open parentheses 2 alpha plus beta close parentheses$ is equal to, $alpha comma blank beta element of open parentheses 0 comma blank pi divided by 2 close parentheses$

$sin invisible function application open parentheses alpha plus beta close parentheses equals 1 comma sin invisible function application open parentheses alpha minus beta close parentheses equals fraction numerator 1 over denominator 2 end fraction comma blank$ then $tan invisible function application open parentheses alpha plus 2 beta close parentheses tan invisible function application open parentheses 2 alpha plus beta close parentheses$ is equal to, $alpha comma blank beta element of open parentheses 0 comma blank pi divided by 2 close parentheses$

$alpha plus beta plus gamma equals 2 pi comma$ then

$alpha plus beta plus gamma equals 2 pi comma$ then

parallel

$i s blank e q u a l blank t o$

$i s blank e q u a l blank t o$

t area of triangle $A B C$ is $blank open parentheses square root of 3 minus 1 close parentheses divided by 2 comma blank b equals 2$ and $blank c equals open parentheses square root of 3 minus 1 close parentheses$ and $blank angle A$ is acute. The measure of the angle $C$ is

t area of triangle $A B C$ is $blank open parentheses square root of 3 minus 1 close parentheses divided by 2 comma blank b equals 2$ and $blank c equals open parentheses square root of 3 minus 1 close parentheses$ and $blank angle A$ is acute. The measure of the angle $C$ is

$tan invisible function application beta equals 2 sin invisible function application alpha sin invisible function application gamma blank c o s e c blank open parentheses alpha plus gamma close parentheses comma blank t h e n cot invisible function application alpha comma cot invisible function application beta comma cot invisible function application gamma blank$ are in

$tan invisible function application beta equals 2 sin invisible function application alpha sin invisible function application gamma blank c o s e c blank open parentheses alpha plus gamma close parentheses comma blank t h e n cot invisible function application alpha comma cot invisible function application beta comma cot invisible function application gamma blank$ are in

parallel

e equation $sin to the power of 2 end exponent invisible function application theta equals fraction numerator x to the power of 2 end exponent plus y to the power of 2 end exponent over denominator 2 x y end fraction$ is possible if

e equation $sin to the power of 2 end exponent invisible function application theta equals fraction numerator x to the power of 2 end exponent plus y to the power of 2 end exponent over denominator 2 x y end fraction$ is possible if

$increment A B C comma sin invisible function application A plus sin invisible function application B plus sin invisible function application C equals 1 plus square root of 2$ and $cos invisible function application A plus cos invisible function application plus cos invisible function application C equals square root of 2$ if the triangle is

$increment A B C comma sin invisible function application A plus sin invisible function application B plus sin invisible function application C equals 1 plus square root of 2$ and $cos invisible function application A plus cos invisible function application plus cos invisible function application C equals square root of 2$ if the triangle is

ven $A equals sin to the power of 2 end exponent invisible function application theta plus cos to the power of 2 end exponent invisible function application theta comma$ then for all real $theta$ ,

ven $A equals sin to the power of 2 end exponent invisible function application theta plus cos to the power of 2 end exponent invisible function application theta comma$ then for all real $theta$ ,

parallel

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