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Maths-

General

Easy

Question

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

$\tan \frac{α}{2} + \tan \frac{β}{2}$ + $\tan \frac{γ}{2} = \tan \frac{α}{2} \tan \frac{β}{2} \tan \frac{γ}{2}$
$\tan \frac{α}{2} \tan \frac{β}{2} + \tan \frac{β}{2} \tan \frac{γ}{2}$ + $\tan \frac{γ}{2} \tan \frac{α}{2} = 1$
$\tan \frac{α}{2} + \tan \frac{β}{2}$ + $\tan \frac{γ}{2} = - \tan \frac{α}{2} \tan \frac{β}{2} \tan \frac{γ}{2}$
None of these

The correct answer is: $tan invisible function application fraction numerator alpha over denominator 2 end fraction plus tan invisible function application fraction numerator beta over denominator 2 end fraction$ + $tan invisible function application fraction numerator gamma over denominator 2 end fraction equals tan invisible function application fraction numerator alpha over denominator 2 end fraction tan invisible function application fraction numerator beta over denominator 2 end fraction tan invisible function application fraction numerator gamma over denominator 2 end fraction$

$alpha plus beta plus gamma equals 2 pi rightwards double arrow fraction numerator alpha over denominator 2 end fraction plus fraction numerator beta over denominator 2 end fraction plus fraction numerator gamma over denominator 2 end fraction equals pi$
$therefore tan invisible function application open parentheses fraction numerator alpha over denominator 2 end fraction plus fraction numerator beta over denominator 2 end fraction close parentheses equals tan invisible function application open parentheses pi minus fraction numerator gamma over denominator 2 end fraction close parentheses equals negative tan invisible function application fraction numerator gamma over denominator 2 end fraction$
$rightwards double arrow fraction numerator tan invisible function application alpha divided by 2 plus tan invisible function application beta divided by 2 over denominator 1 minus tan invisible function application alpha divided by 2 tan invisible function application beta divided by 2 end fraction equals tan invisible function application gamma divided by 2$
$rightwards double arrow tan invisible function application alpha divided by 2 plus tan invisible function application beta divided by 2 plus tan invisible function application gamma divided by 2 equals tan invisible function application alpha divided by 2 tan invisible function application beta divided by 2 tan invisible function application gamma divided by 2$ $left square bracket because cos to the power of 2 end exponent invisible function application theta less or equal than 1 right square bracket$

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

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

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

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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 $theta$ ,

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

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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

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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

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