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Question

If $open vertical bar 2 sin invisible function application theta minus c o s e c blank theta close vertical bar greater or equal than 1$ and $theta not equal to fraction numerator n pi over denominator 2 end fraction comma blank n element of I$ , then

$\cos 2 θ \geq \frac{1}{2}$
$\cos 2 θ \geq \frac{1}{4}$
$\cos 2 θ \leq \frac{1}{2}$
$\cos 2 θ \leq 1 / 4$

The correct answer is: $cos invisible function application 2 theta greater or equal than fraction numerator 1 over denominator 2 end fraction$

$open vertical bar 2 sin invisible function application theta minus blank c o s e c blank theta close vertical bar greater or equal than 1$
$rightwards double arrow open vertical bar 2 sin to the power of 2 end exponent invisible function application theta minus 1 close vertical bar greater or equal than open vertical bar sin invisible function application theta close vertical bar$
$rightwards double arrow open vertical bar cos invisible function application 2 theta close vertical bar greater or equal than open vertical bar sin invisible function application theta close vertical bar$
$rightwards double arrow 2 cos to the power of 2 end exponent invisible function application 2 theta greater or equal than 1 minus cos invisible function application 2 theta$
$rightwards double arrow 2 cos to the power of 2 end exponent invisible function application 2 theta plus cos invisible function application 2 theta minus 1 greater or equal than 0$
$rightwards double arrow open parentheses 2 cos invisible function application 2 theta minus 1 close parentheses open parentheses cos invisible function application 2 theta plus 1 close parentheses greater or equal than 0$
$rightwards double arrow cos invisible function application 2 theta greater or equal than fraction numerator 1 over denominator 2 end fraction blank open square brackets a s cos invisible function application theta not equal to 0 comma blank i. e. comma cos invisible function application 2 theta not equal to negative 1 close square brackets$

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If $S equals open curly brackets x element of N blank colon 2 plus log subscript 2 end subscript invisible function application square root of x plus 1 end root greater than 1 minus log subscript 1 divided by 2 end subscript invisible function application square root of 4 minus x to the power of 2 end exponent end root close curly brackets comma$ then

If $S equals open curly brackets x element of N blank colon 2 plus log subscript 2 end subscript invisible function application square root of x plus 1 end root greater than 1 minus log subscript 1 divided by 2 end subscript invisible function application square root of 4 minus x to the power of 2 end exponent end root close curly brackets comma$ then

$log subscript 4 end subscript invisible function application 18$ is

$log subscript 4 end subscript invisible function application 18$ is

If $sin invisible function application 6 theta plus sin invisible function application 4 theta plus sin invisible function application 2 theta equals 0 comma$ then $theta blank$ is equal to $blank open parentheses n element of Z close parentheses$

If $sin invisible function application 6 theta plus sin invisible function application 4 theta plus sin invisible function application 2 theta equals 0 comma$ then $theta blank$ is equal to $blank open parentheses n element of Z close parentheses$

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The area of the circle and the area of a regular polygon of $n$ sides and of perimeter equal to that of the circle are in the ratio of

The area of the circle and the area of a regular polygon of $n$ sides and of perimeter equal to that of the circle are in the ratio of

$I f blank x element of open parentheses pi comma fraction numerator 3 pi over denominator 2 end fraction close parentheses comma blank t h e n blank 4 cos to the power of 2 end exponent invisible function application open parentheses fraction numerator pi over denominator 4 end fraction minus fraction numerator x over denominator 2 end fraction close parentheses plus square root of 4 sin to the power of 4 end exponent invisible function application x plus sin to the power of 2 end exponent invisible function application 2 x end root$ is always equal to

$I f blank x element of open parentheses pi comma fraction numerator 3 pi over denominator 2 end fraction close parentheses comma blank t h e n blank 4 cos to the power of 2 end exponent invisible function application open parentheses fraction numerator pi over denominator 4 end fraction minus fraction numerator x over denominator 2 end fraction close parentheses plus square root of 4 sin to the power of 4 end exponent invisible function application x plus sin to the power of 2 end exponent invisible function application 2 x end root$ is always equal to

If $tan invisible function application theta plus sin invisible function application theta equals m blank a n d tan invisible function application theta minus sin invisible function application theta equals n comma$ then

If $tan invisible function application theta plus sin invisible function application theta equals m blank a n d tan invisible function application theta minus sin invisible function application theta equals n comma$ then

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$I f blank x subscript 1 end subscript blank a n d blank x subscript 2 end subscript$ are two distinct roots of the equation $a cos invisible function application x plus b sin invisible function application x equals c comma blank t h e n tan invisible function application fraction numerator x subscript 1 end subscript plus x subscript 2 end subscript over denominator 2 end fraction$ is equal to

$I f blank x subscript 1 end subscript blank a n d blank x subscript 2 end subscript$ are two distinct roots of the equation $a cos invisible function application x plus b sin invisible function application x equals c comma blank t h e n tan invisible function application fraction numerator x subscript 1 end subscript plus x subscript 2 end subscript over denominator 2 end fraction$ is equal to

$I f fraction numerator sin invisible function application open parentheses x plus y close parentheses over denominator sin invisible function application open parentheses x minus y close parentheses end fraction equals fraction numerator a plus b over denominator a minus b end fraction comma blank t h e n fraction numerator tan invisible function application x over denominator tan invisible function application y end fraction blank i s blank e q u a l blank t o$

$I f fraction numerator sin invisible function application open parentheses x plus y close parentheses over denominator sin invisible function application open parentheses x minus y close parentheses end fraction equals fraction numerator a plus b over denominator a minus b end fraction comma blank t h e n fraction numerator tan invisible function application x over denominator tan invisible function application y end fraction blank i s blank e q u a l blank t o$

Which of the following is not the general solution of $2 to the power of cos invisible function application 2 x end exponent plus 1 equals 3.2 to the power of negative sin to the power of 2 end exponent invisible function application x end exponent ?$

Which of the following is not the general solution of $2 to the power of cos invisible function application 2 x end exponent plus 1 equals 3.2 to the power of negative sin to the power of 2 end exponent invisible function application x end exponent ?$

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In a transformer, number of turns in the primary are 140 and that in the secondary are 280. If current in primary is $4 A$ , then that in the secondary is

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A transformer is often filled with oil. The oil used should have

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