Question

show that 1+tan^2A/cot^2A=sec^4A-sec^2A

anyone answer this sum​

Answer 1

Answer:

taking lhs

Sec²/cot²

sin²

taking rhs

sec²(sec²-1)

sec²(tan) ²

sin²

hence proved

Answer 2

To Prove:

$\dfrac{1 + {\tan}^{2}A }{{\cot}^{2}A} = {\sec }^{4}A - {\sec}^{2}A$

Solution:

L.H.S

$=\dfrac{1 + {\tan}^{2} A}{{\cot}^{2}A} \\ \\ [\text{Using(1)}]\\ \\ = \frac{ { \sec }^{2}A }{ \cot {}^{2} A} \\ \\ [\text{Using(2)}]\\ \\ = { \sec}^{2}A. { \tan }^{2}A \\ \\ = { \sec}^{2} A( { \sec}^{2} A - 1) \\ \\ = {\sec}^{4}A - {\sec}^{2}A$

$\therefore$ L.H.S = R.H.S [Proved]

Used Identity and Rule

sec²@ - tan²@ = 1 ---(1)

tan @ = 1/cot @ ---(2)

Important Trigonometry Identities

sin²@ + cos²@ = 1

cosec²@- cot²@ = 1

Basic Trigonometry Rule

sin @ = 1 /cosec @

cos @ = 1 /sec @

tan @ = 1 /cot @ = sin @/cos @