Question

cosec^2Acot^2A - sec^2Atan^2A - (cot^2A-tan^2A)(sec^2Acosec^2A-1) = ?​

Answer 1

Answer: 0

Step-by-step explanation:

$cosec^{2}A.cot^{2}A-sec^{2}A.tan^{2}A-(cot^{2}A-tan^{2}A)(sec^{2}A.cosec^{2}A-1)$

We know that cosec A = 1/sin A ,cot A = cos A/sin A, sec A= 1/cos A and tan A =sin A/ cos A.

$\frac{1}{sin^{2}A } .\frac{cos^{2}A }{sin^{2}A }-\frac{1}{cos^{2}A }.\frac{sin^{2}A }{cos^{2}A } -(\frac{cos^{2}A }{sin^{2}A } -\frac{sin^{2}A }{cos^{2}A })(\frac{1}{cos^{2} }.\frac{1}{sin^{2}A }-1)$

$\frac{cos^{2}A }{sin^{4}A }-\frac{sin^{2}A }{cos^{4}A }-(\frac{cos^{4}A-sin^{4}A }{sin^{2}A.cos^{2}A } )(\frac{1-cos^{2}A.sin^{2}A }{cos^{2}A.sin^{2} A } )$

$\frac{cos^{6}A-sin^{6}A }{sin^{4}.cos^{4}A }-(\frac{cos^{4}A-sin^{4}A-cos^{6}A.sin^{2}A+cos^{2}A.sin^{6}A }{cos^{4}A.sin^{4}A } )$$\frac{(cos^{2}A-sin^{2}A)(1+sin^{2}A.cos^{2}A) }{sin^{4}A.cos^{4}A } -[\frac{(cos^{2}A-sin^{2}A)-cos^{2}A.sin^{2}A(sin^{2}A-cos^{2}A) }{cos^{2}A.sin^{2}A } ]$

$\frac{cos^{2}A-sin^{2}A }{sin^{4}A.cos^{4}A } [(1+sin^{2}A.cos^{2}A)- [(1+sin^{2}A.cos^{2}A)$

$\frac{cos^{2}A-sin^{2}A }{sin^{4}A.cos^{4}A }(0)\\=0$

Answer 2

ANSWER:

ZERO

TO FIND:

cosec² A cot² A - sec² A tan² A - (cot² A - tan² A)(sec² A cosec² A - 1)

FORMULAE:

sin² ∅ + cos² ∅ = 1

cosec ∅ = 1 / sin ∅

sec ∅ = 1 / cos ∅

tan ∅ = sin ∅ / cos ∅

cot ∅ = cos ∅ / sin ∅

(A³ - B³) = (A - B)(A² + B² + AB)

(A² + B²) = (A + B)² - 2AB

EXPLANATION:

Take cosec² A cot² A - sec² A tan² A

$\displaystyle \frac{1}{ { \sin}^{2} A} \times \frac{ { \cos}^{2}A }{ { \sin}^{2}A } - \frac{1}{ { \cos}^{2} A} \times \frac{ { \sin}^{2}A }{ { \cos}^{2}A }\\ \\ \\ \frac{ { \cos}^{2}A }{ { \sin}^{4}A} - \frac{ { \sin}^{2}A }{ { \cos}^{4}A } \\ \\ \\ \frac{ { \cos}^{6}A - {\sin}^{6}A}{ { \sin}^{4}A \: { \cos}^{4}A } \\ \\ \\ \frac{ { {( \cos}^{2} A)}^{3} - { {( \sin}^{2} A)}^{3} }{ { \sin}^{4}A \: { \cos}^{4}A} \\ \\ \\ \frac{ ({ \cos}^{2}A- { \sin}^{2} A)( { \cos}^{4}A+ { \sin}^{4} A + sinA \: \cos A ) }{{ \sin}^{4}A\: { \cos}^{4}A} \\ \\ \\ \frac{ ({ \cos}^{2}A - { \sin}^{2} A)( {( { \cos}^{2}A + { \sin}^{2} A )}^{2} - 2 \: sinA \: \cos A+ sinA \: \cos A ) }{{ \sin}^{4}A \: { \cos}^{4}A} \\ \\ \frac{{ \cos}^{2} A- { \sin}^{2}A(1 - { \sin}^{2}A\: { \cos}^{2} A)}{ { \sin}^{4}A \: { \cos}^{4}A}$

Take -(cot² A - tan² A)(sec² A cosec² A - 1)

$\displaystyle - \left( \frac{ { \cos}^{2}a }{ { \sin}^{2}A} - \frac{ {\sin}^{2}A }{ { \cos}^{2}A}\right) \left( \frac{1}{ { \cos}^{2} A } \times \frac{1}{ { \sin}^{2}A } - 1 \right) \\ \\ - \left( \frac{ { \cos}^{4} A - { \sin}^{4} A}{{ \sin}^{2}A \: { \cos}^{2} A} \right) \left( \frac{1 - { \sin}^{2}A \: { \cos}^{2}A}{{ \sin}^{2}A\: { \cos}^{2}A} \right) \\ \\ { \cos}^{4}A- { \sin}^{4} A= ({cos}^{2}A+ { \sin}^{2}A) ({cos}^{2} A - { \sin}^{2}A) \\ \\ { \cos}^{4} A - { \sin}^{4} A= { \cos}^{2} A- { \sin}^{2}A \\ \\ - \frac{{ \cos}^{2} A- { \sin}^{2}A(1 - { \sin}^{2}A\: { \cos}^{2} A)}{ { \sin}^{4}A \: { \cos}^{4}A}$

ADDING THE TWO TERMS WE WILL GET ZERO AS BOTH THE TERMS ARE SAME AND POSSESS A NEGATIVE SYMBOL BETWEEN THEM.