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Maths-

General

Easy

Question

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

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

The correct answer is: A.P.

$tan invisible function application beta equals 2 sin invisible function application alpha sin invisible function application gamma blank c o s e c blank open parentheses alpha plus gamma close parentheses equals fraction numerator 2 sin invisible function application alpha sin invisible function application gamma over denominator sin invisible function application open parentheses alpha plus gamma close parentheses end fraction$
$rightwards double arrow cot invisible function application beta equals fraction numerator sin invisible function application open parentheses alpha plus gamma close parentheses over denominator 2 sin invisible function application alpha sin invisible function application gamma end fraction$
$rightwards double arrow 2 cot invisible function application beta equals fraction numerator sin invisible function application alpha cos invisible function application gamma plus cos invisible function application alpha sin invisible function application gamma over denominator sin invisible function application alpha sin invisible function application gamma end fraction equals cot invisible function application alpha plus cot invisible function application gamma$
$rightwards double arrow cot invisible function application alpha comma cot invisible function application beta comma cot invisible function application gamma blank a r e blank i n blank A. P.$

Related Questions to study

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

$open parentheses 1 plus tan invisible function application alpha close parentheses open parentheses 1 plus tan invisible function application 4 alpha close parentheses equals 2 comma blank alpha element of open parentheses 0 comma blank pi divided by 16 close parentheses$ then $alpha$ is equal to

$open parentheses 1 plus tan invisible function application alpha close parentheses open parentheses 1 plus tan invisible function application 4 alpha close parentheses equals 2 comma blank alpha element of open parentheses 0 comma blank pi divided by 16 close parentheses$ then $alpha$ is equal to

Let $f open parentheses theta close parentheses equals sin invisible function application theta left parenthesis sin invisible function application theta plus sin invisible function application 3 theta right parenthesis.$ Then $f left parenthesis theta right parenthesis$ is

Let $f open parentheses theta close parentheses equals sin invisible function application theta left parenthesis sin invisible function application theta plus sin invisible function application 3 theta right parenthesis.$ Then $f left parenthesis theta right parenthesis$ is

The value of $tan invisible function application 9 degree minus tan invisible function application 27 degree minus tan invisible function application 63 degree plus tan invisible function application 81 degree blank i s$

The value of $tan invisible function application 9 degree minus tan invisible function application 27 degree minus tan invisible function application 63 degree plus tan invisible function application 81 degree blank i s$

parallel

If $blank sin invisible function application x plus cos invisible function application x equals fraction numerator square root of 7 over denominator 2 end fraction blank$ where $blank x element of A comma$ then $tan invisible function application fraction numerator x over denominator 2 end fraction blank$ is equal to

If $blank sin invisible function application x plus cos invisible function application x equals fraction numerator square root of 7 over denominator 2 end fraction blank$ where $blank x element of A comma$ then $tan invisible function application fraction numerator x over denominator 2 end fraction blank$ is equal to

If $fraction numerator 3 pi over denominator 4 end fraction less than alpha less than pi comma blank$ then $square root of 2 cot invisible function application alpha plus fraction numerator 1 over denominator sin to the power of 2 end exponent invisible function application alpha end fraction end root blank$ is equal to

If $fraction numerator 3 pi over denominator 4 end fraction less than alpha less than pi comma blank$ then $square root of 2 cot invisible function application alpha plus fraction numerator 1 over denominator sin to the power of 2 end exponent invisible function application alpha end fraction end root blank$ is equal to

Hydrogen burns in air with a

Hydrogen burns in air with a

Chemistry-General

parallel

Metal hydride on treatment with water gives

Metal hydride on treatment with water gives

Chemistry-General

Which is used hydrogen generators

Which is used hydrogen generators

Chemistry-General

Assertion:First ionization energy for nitrogen is lower than oxygen.
Reason :Across a period effective nuclear charge decreases.

Assertion:First ionization energy for nitrogen is lower than oxygen.
Reason :Across a period effective nuclear charge decreases.

Chemistry-General

parallel

Assertion :Outermost electronic configuration of most electropositive elements is ns² np³.
Reason :ns² np³is stable due to half filled subshell.

Assertion :Outermost electronic configuration of most electropositive elements is ns² np³.
Reason :ns² np³is stable due to half filled subshell.

Physics-General

Assertion :NO^– ion is isoelectronic with CN^– ion.
Reason :Isoelectronic ions have same number of elelctrons.

Assertion :NO^– ion is isoelectronic with CN^– ion.
Reason :Isoelectronic ions have same number of elelctrons.

Physics-General

Assertion :I.E. of ₇N is more than that of ₈O as well as ₆C.
Reason : This is due to difference in reactivity towards oxygen.

Assertion :I.E. of ₇N is more than that of ₈O as well as ₆C.
Reason : This is due to difference in reactivity towards oxygen.

Physics-General

parallel

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