If mcos2A-nsin2A-p- then 2A-

mcos ^ 2 A +nsin ^ 2 A =p ⇒ m( 1 sin ^ 2 A #160; ) +nsin ^ 2 A =p ⇒ m+( n m ) sin ^ 2 A =p ⇒sin ^ 2 A = p m / n m .......( ...

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Standard XII

Mathematics

Basic Trigonometric Identities

If mcos2A+n...

Question

If $m {cos}^{2} A + n {sin}^{2} A = p$ , then ${cot}^{2} A =$

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Solution

$m {cos}^{2} A + n {sin}^{2} A = p \Rightarrow m (1 - {sin}^{2} A) + n {sin}^{2} A = p \Rightarrow m + (n - m) {sin}^{2} A = p \Rightarrow {sin}^{2} A = \frac{p - m}{n - m} . . . . . . . (1) \Rightarrow 1 - {cos}^{2} A = \frac{p - m}{n - m} \Rightarrow {cos}^{2} A = 1 - \frac{p - m}{n - m} \Rightarrow {cos}^{2} A = \frac{n - m - p + m}{n - m} \Rightarrow {cos}^{2} A = \frac{n - p}{n - m} . . . . . (2)$
from $(1)$ and $(2)$ ${cot}^{2} A = \frac{{cos}^{2} A}{{sin}^{2} A} \Rightarrow {cot}^{2} A = \frac{\frac{n - p}{n - m}}{\frac{p - m}{n - m}} \Rightarrow {cot}^{2} A = \frac{n - p}{p - m}$

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If mcos2A+nsin2A=p, then cot2A=

mcos2A+nsin2A=p⇒m(1−sin2A)+nsin2A=p⇒m+(n−m)sin2A=p⇒sin2A=p−mn−m.......(1)⇒1−cos2A=p−mn−m⇒cos2A=1−p−mn−m⇒cos2A=n−m−p+mn−m⇒cos2A=n−pn−m.....(2)from (1) and (2) cot2A=cos2Asin2A⇒cot2A=n−pn−mp−mn−m⇒cot2A=n−pp−m

If $m {cos}^{2} A + n {sin}^{2} A = p$ , then ${cot}^{2} A =$