Statement 1:The redox titrations in which liberated $straight I subscript 2$ is used as oxidant are called as iodometric titrations
Statement 2:Addition of KI of $CuSO subscript 4$ liberates $straight I subscript 2$ which is estimated against hypo solution.

Statement-I : If $sin squared space A equals sin squared space B$ and $cos squared space A equals cos squared space B$ then $straight A equals straight n pi plus straight B comma space straight n element of straight I$
Statement-II : If sinA = sinB and cosA = cosB, then $A equals n pi plus B comma n element of I$

In this question, we have to find the statements are the correct or not and statement 2 is correct explanation or not, is same as assertion and reason. Here, Start solving first Statement and try to prove it. Then solve the Statement-II.

Statement-I : If $sin squared space A equals sin squared space B$ and $cos squared space A equals cos squared space B$ then $straight A equals straight n pi plus straight B comma space straight n element of straight I$
Statement-II : If sinA = sinB and cosA = cosB, then $A equals n pi plus B comma n element of I$

chemistry-

Arsine is:

chemistry-General

maths-

Statement-I : The equation sin(cos x) = cos(sin x) does not possess real roots.
Statement-II : If sin x > 0, then $2 n pi less than x less than left parenthesis 2 n plus 1 right parenthesis comma n element of I$

maths-General

Statement-I : In (0, $straight pi$ ), the number of solutions of the equation $tan space theta plus tan space 2 theta plus tan space 3 theta equals tan space theta tan space 2 theta tan space 3 theta$ is two
Statement-II : $tan space 6 theta$ is not defined at $theta equals left parenthesis 2 straight n plus 1 right parenthesis pi over 12 comma straight n element of straight I$

In this question, we have to find the statements are the correct or not and statement 2 is correct explanation or not, is same as assertion and reason. Here, it has 5 solutions, but tanθ &tan3θ are not defined at $straight pi over 2$ , $straight pi over 6$ , $fraction numerator 5 straight pi over denominator 6 end fraction$ . respectively so it remains only 2.

Statement-I : In (0, $straight pi$ ), the number of solutions of the equation $tan space theta plus tan space 2 theta plus tan space 3 theta equals tan space theta tan space 2 theta tan space 3 theta$ is two
Statement-II : $tan space 6 theta$ is not defined at $theta equals left parenthesis 2 straight n plus 1 right parenthesis pi over 12 comma straight n element of straight I$

Statement-I : If sin x + cos x = $square root of open parentheses y plus 1 over y close parentheses end root comma x element of left square bracket 0 comma pi right square bracket$ then $x equals pi over 4 comma y equals 1$
Statement-II : AM ≥ GM

Statement-I : The number of real solutions of the equation sin x = 2^x + 2^–x is zero
Statement-II : Since |sin x| ≤ 1

In this question, we have to find the statements are the correct or not and statement 2 is correct explanation or not, is same as assertion and reason. Here, -1 ≤ sinx ≤ 1 for all value, remember that.

Statement-I : The number of real solutions of the equation sin x = 2^x + 2^–x is zero
Statement-II : Since |sin x| ≤ 1

maths-

$sin space x plus sin space 2 x plus sin space 3 x equals cos space x plus cos space 2 x plus cos space 3 x$ if

$4 sin to the power of 4 space x plus cos to the power of 4 space x equals 1$ if

In this question, we have to find the general solution of x. Here more than one option will correct. Remember the rules for finding the general solution.

$4 sin to the power of 4 space x plus cos to the power of 4 space x equals 1$ if

In this question, we have to find the general solution of x. Here more than one option will correct. Remember the rules for finding the general solution.

Number of ordered pairs (a, x) satisfying the equation $sec squared space left parenthesis a plus 2 right parenthesis x plus a squared minus 1 equals 0 semicolon minus pi less than x less than pi$ is

The general solution of the equation, $fraction numerator 1 minus sin space x plus midline horizontal ellipsis comma plus left parenthesis negative 1 right parenthesis to the power of n sin to the power of n space x plus midline horizontal ellipsis straight infinity over denominator 1 plus sin space x plus midline horizontal ellipsis plus sin to the power of n space x plus midline horizontal ellipsis straight infinity end fraction equals fraction numerator 1 minus cos space 2 x over denominator 1 plus cos space 2 x end fraction$ is :