Question

Prove that cot^2A-cot^2B=sin^2B-sin^2A/sin^2A x sin^2B

Answer 1

${ \cot }^{2} a - { \cot}^{2} b \\ { \csc }^{2} a - 1 - ( { \csc }^{2} b - 1) \\ \frac{1}{ { \sin}^{2}a } - \frac{1}{ { \sin }^{2}b } \\ { \sin}^{2} b - { \sin}^{2} a \div sin \: a \times sin \: b$
this is the verification

Answer 2

Concept:

1+cot²A=cosec²A

1/sinA=cosecA

Given:

LHS=cot²A-cot²b

Find:

LHS=RHS=sin²B-sin²A/sin²A x sin²B

Solution:

LHS=cot²A-cot²B

      = csc²A-1-csc²B+1

      =1/sin²A-1/Sin²B

       =sin²B-sin²A/sin²A x sin²B=RHS

Hence proved

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