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Maths-

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Question

Assertion A: In a right-angled triangle $s i n squared space A plus s i n squared space B plus s e c squared space C equals 2$ Reason $R colon$ if $alpha comma beta$ are complementary angles then $s i n squared space alpha plus s i n squared space beta equals 1$

A, R are true and R is the correct explanation of A
A, R are true and R is not the correct explanation of A
A is true, R is false
A is false, R is true

The correct answer is: A is false, R is true

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AB is a line segment of length 24 cm and C is its middle point. On AB, A and CB semi-circles are described. The radius of circle which touches all these semi-circles is

AB is a line segment of length 24 cm and C is its middle point. On AB, A and CB semi-circles are described. The radius of circle which touches all these semi-circles is

in a $triangle P Q R$ $3 sin space P plus 4 cos space Q equals 6$ and $4 sin space Q plus 3 cos$ p=1 then

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The expression $fraction numerator tan space A over denominator 1 minus cat space A end fraction plus fraction numerator cot space A over denominator 1 minus tan space d end fraction$ can be written as

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if $straight f subscript straight k left parenthesis straight x right parenthesis equals 1 divided by straight k left parenthesis sin to the power of straight k space straight x plus cos to the power of straight k space straight x right parenthesis$ where $straight x element of straight R comma straight k greater or equal than 1$ then $straight f subscript 4 left parenthesis straight x right parenthesis minus straight f subscript 6 left parenthesis straight x right parenthesis equals$

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$integral fraction numerator d x over denominator square root of x squared plus 2 x plus 1 end root end fraction$ = A log |x + 1| + C for x < – 1 then A is -

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