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Question

If $alpha comma blank beta comma blank gamma comma blank delta$ are the smallest positive angles in ascending order of magnitude which have their sines equal to the positive quantity $k comma$ then the value of $4 sin invisible function application fraction numerator alpha over denominator 2 end fraction plus 3 sin invisible function application fraction numerator beta over denominator 2 end fraction plus 2 sin invisible function application fraction numerator gamma over denominator 2 end fraction plus sin invisible function application fraction numerator delta over denominator 2 end fraction$ is equal to

$2 \sqrt{1 - k}$
$2 \sqrt{1 + k}$
$\frac{\sqrt{1 + k}}{2}$
None of these

The correct answer is: $2 square root of 1 plus k end root$

Since $blank alpha less than beta less than gamma less than delta$ and $sin invisible function application alpha equals sin invisible function application beta equals sin invisible function application gamma equals sin invisible function application delta equals K comma blank$ therefore $blank beta equals pi minus alpha comma blank gamma$
$equals 2 pi plus alpha comma blank delta equals 3 pi minus alpha$
$rightwards double arrow 4 sin invisible function application fraction numerator alpha over denominator 2 end fraction plus 3 sin invisible function application fraction numerator beta over denominator 2 end fraction plus 2 sin invisible function application fraction numerator gamma over denominator 2 end fraction plus sin invisible function application fraction numerator delta over denominator 2 end fraction equals 4 sin invisible function application fraction numerator alpha over denominator 2 end fraction plus 3 cos invisible function application fraction numerator alpha over denominator 2 end fraction$
$negative 2 sin invisible function application fraction numerator alpha over denominator 2 end fraction minus cos invisible function application fraction numerator alpha over denominator 2 end fraction$
$equals 2 sin invisible function application fraction numerator alpha over denominator 2 end fraction plus 2 cos invisible function application fraction numerator alpha over denominator 2 end fraction equals 2 square root of 1 plus sin invisible function application alpha end root equals 2 square root of 1 plus K end root$

Related Questions to study

The value of $x$ satisfying $square root of 3 to the power of negative 4 plus 2 log subscript square root of 5 end subscript invisible function application x end exponent equals 1 divided by 9$ is

The value of $x$ satisfying $square root of 3 to the power of negative 4 plus 2 log subscript square root of 5 end subscript invisible function application x end exponent equals 1 divided by 9$ is

In $blank increment A B C$ , if $b to the power of 2 end exponent plus c to the power of 2 end exponent equals 2 a to the power of 2 end exponent comma$ then value of $fraction numerator cot invisible function application A over denominator cot invisible function application B plus cot invisible function application C end fraction$ is

In $blank increment A B C$ , if $b to the power of 2 end exponent plus c to the power of 2 end exponent equals 2 a to the power of 2 end exponent comma$ then value of $fraction numerator cot invisible function application A over denominator cot invisible function application B plus cot invisible function application C end fraction$ is

If $log subscript 10 end subscript invisible function application open square brackets fraction numerator 1 over denominator 2 to the power of x end exponent plus blank x minus 1 end fraction close square brackets equals x open square brackets log subscript 10 end subscript invisible function application 5 minus 1 close square brackets comma blank t h e n blank x equals$

If $log subscript 10 end subscript invisible function application open square brackets fraction numerator 1 over denominator 2 to the power of x end exponent plus blank x minus 1 end fraction close square brackets equals x open square brackets log subscript 10 end subscript invisible function application 5 minus 1 close square brackets comma blank t h e n blank x equals$

parallel

r triangle $A B C comma blank R equals 5 divided by 2 blank$ and $blank r equals 1.$ Let $I$ be the incentre of the triangle and $D comma blank E$ and $blank F$ be the feet of the perpendiculars from $I blank$ to $blank B C comma blank C A blank$ and $blank A B comma$ respectively. The value of $fraction numerator I D cross times I E cross times I F over denominator I A cross times I B cross times I C end fraction blank$ is equal to

r triangle $A B C comma blank R equals 5 divided by 2 blank$ and $blank r equals 1.$ Let $I$ be the incentre of the triangle and $D comma blank E$ and $blank F$ be the feet of the perpendiculars from $I blank$ to $blank B C comma blank C A blank$ and $blank A B comma$ respectively. The value of $fraction numerator I D cross times I E cross times I F over denominator I A cross times I B cross times I C end fraction blank$ is equal to

any $increment A B C comma$ the value of $a open parentheses b to the power of 2 end exponent plus c to the power of 2 end exponent close parentheses cos invisible function application A plus b left parenthesis c to the power of 2 end exponent plus a to the power of 2 end exponent right parenthesis cos invisible function application B plus c left parenthesis a to the power of 2 end exponent plus b to the power of 2 end exponent right parenthesis cos invisible function application C equals$

any $increment A B C comma$ the value of $a open parentheses b to the power of 2 end exponent plus c to the power of 2 end exponent close parentheses cos invisible function application A plus b left parenthesis c to the power of 2 end exponent plus a to the power of 2 end exponent right parenthesis cos invisible function application B plus c left parenthesis a to the power of 2 end exponent plus b to the power of 2 end exponent right parenthesis cos invisible function application C equals$

$fraction numerator sin invisible function application x over denominator sin invisible function application y end fraction equals fraction numerator 1 over denominator 2 end fraction comma fraction numerator cos invisible function application x over denominator cos invisible function application y end fraction equals fraction numerator 3 over denominator 2 end fraction blank w h e r e blank x comma blank y blank element of open parentheses 0 comma fraction numerator pi over denominator 2 end fraction close parentheses comma blank t h e n blank t h e blank v a l u e blank o f blank left parenthesis x plus y right parenthesis blank i s blank e q u a l blank t o$

$fraction numerator sin invisible function application x over denominator sin invisible function application y end fraction equals fraction numerator 1 over denominator 2 end fraction comma fraction numerator cos invisible function application x over denominator cos invisible function application y end fraction equals fraction numerator 3 over denominator 2 end fraction blank w h e r e blank x comma blank y blank element of open parentheses 0 comma fraction numerator pi over denominator 2 end fraction close parentheses comma blank t h e n blank t h e blank v a l u e blank o f blank left parenthesis x plus y right parenthesis blank i s blank e q u a l blank t o$

parallel

$tan invisible function application 3 theta plus tan invisible function application theta equals 2 tan invisible function application 2 theta$ , then $blank theta$ is equal to $open parentheses n element of Z close parentheses$

$tan invisible function application 3 theta plus tan invisible function application theta equals 2 tan invisible function application 2 theta$ , then $blank theta$ is equal to $open parentheses n element of Z close parentheses$

e number of solutions of $blank not stretchy sum from r equals 1 to 5 of cos invisible function application r x equals 5$ in the interval $blank open square brackets 0 comma blank 2 pi close square brackets blank$ is

e number of solutions of $blank not stretchy sum from r equals 1 to 5 of cos invisible function application r x equals 5$ in the interval $blank open square brackets 0 comma blank 2 pi close square brackets blank$ is

Member of roots of $open parentheses 1 minus tan invisible function application theta close parentheses open parentheses 1 plus sin invisible function application 2 theta close parentheses equals 1 plus tan invisible function application theta$ for $blank theta element of open square brackets 0 comma blank 2 pi close square brackets$ is

Member of roots of $open parentheses 1 minus tan invisible function application theta close parentheses open parentheses 1 plus sin invisible function application 2 theta close parentheses equals 1 plus tan invisible function application theta$ for $blank theta element of open square brackets 0 comma blank 2 pi close square brackets$ is

parallel

r $blank n element of Z comma blank$ the general solution of $open parentheses square root of 3 minus 1 close parentheses sin invisible function application theta plus open parentheses square root of 3 plus 1 close parentheses cos invisible function application theta equals 2 blank$ is $open parentheses n element of Z close parentheses$

r $blank n element of Z comma blank$ the general solution of $open parentheses square root of 3 minus 1 close parentheses sin invisible function application theta plus open parentheses square root of 3 plus 1 close parentheses cos invisible function application theta equals 2 blank$ is $open parentheses n element of Z close parentheses$

e set of all $x$ in $open parentheses fraction numerator – pi over denominator 2 end fraction comma fraction numerator pi over denominator 2 end fraction close parentheses blank$ satisfying $blank open vertical bar 4 sin invisible function application x minus 1 close vertical bar less than square root of 5$ is given by

e set of all $x$ in $open parentheses fraction numerator – pi over denominator 2 end fraction comma fraction numerator pi over denominator 2 end fraction close parentheses blank$ satisfying $blank open vertical bar 4 sin invisible function application x minus 1 close vertical bar less than square root of 5$ is given by

$x comma y element of open square brackets 0 comma blank 2 pi close square brackets$ and $sin invisible function application x plus sin invisible function application y equals 2$ , then the value of $blank x plus y$ is

$x comma y element of open square brackets 0 comma blank 2 pi close square brackets$ and $sin invisible function application x plus sin invisible function application y equals 2$ , then the value of $blank x plus y$ is

parallel

e value of $open parentheses 1 plus cos invisible function application fraction numerator pi over denominator 8 end fraction close parentheses open parentheses 1 plus cos invisible function application fraction numerator 3 pi over denominator 8 end fraction close parentheses open parentheses 1 plus cos invisible function application fraction numerator 5 pi over denominator 8 end fraction close parentheses open parentheses 1 plus cos invisible function application fraction numerator 7 pi over denominator 8 end fraction close parentheses$ is

e value of $open parentheses 1 plus cos invisible function application fraction numerator pi over denominator 8 end fraction close parentheses open parentheses 1 plus cos invisible function application fraction numerator 3 pi over denominator 8 end fraction close parentheses open parentheses 1 plus cos invisible function application fraction numerator 5 pi over denominator 8 end fraction close parentheses open parentheses 1 plus cos invisible function application fraction numerator 7 pi over denominator 8 end fraction close parentheses$ is

triangle $A B C comma fraction numerator sin invisible function application A plus sin invisible function application B plus sin invisible function application C over denominator sin invisible function application A plus sin invisible function application B minus sin invisible function application C end fraction$ is equal to

triangle $A B C comma fraction numerator sin invisible function application A plus sin invisible function application B plus sin invisible function application C over denominator sin invisible function application A plus sin invisible function application B minus sin invisible function application C end fraction$ is equal to

e total number of solutions of $ln invisible function application open vertical bar sin invisible function application x close vertical bar equals negative x to the power of 2 end exponent plus 2 x blank$ in $open square brackets negative fraction numerator pi over denominator 2 end fraction comma fraction numerator 3 pi over denominator 2 end fraction close square brackets$ is equal to

e total number of solutions of $ln invisible function application open vertical bar sin invisible function application x close vertical bar equals negative x to the power of 2 end exponent plus 2 x blank$ in $open square brackets negative fraction numerator pi over denominator 2 end fraction comma fraction numerator 3 pi over denominator 2 end fraction close square brackets$ is equal to

parallel

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