Question

Sin^4 theeta - tan^4 theeta = sec^2 theeta + tan^ theeta ( PROVE THIS)

Answer 1

Answer:

$</p><p></p><p> { \sec^{4} θ } - tan^4 θ = sec^2θ + tan^2θ$

$proof..$

$lhs = Sec^4θ - tan^4 θ$

$= ( { \sec}^{2} θ + { \tan}^{2} θ)( { \sec}^{2} θ - { \tan}^{2} θ)$

$= ( { \sec }^{2} θ + { \tan }^{2} θ)$

$hence \: proved.$

Answer 2

Given : -

• 2sinϴ - 1 = 0

To prove : -

• secϴ + tanϴ = √3

Proof :-

We have ,

• => 2sinϴ - 1 = 0
• => 2sinϴ = 1
• => sinϴ = 1/2
• => sinϴ = sin30°
• => ϴ = 30°

Now ,

• => secϴ + tanϴ = sec30° + tan30°
• => secϴ + tanϴ = 2/√3 + 1/√3
• => secϴ + tanϴ = (2 + 1)/√3
• => secϴ + tanϴ = 3/√3
• => secϴ + tanϴ = √3

Hence proved !