Question

sec squared theta minus tan squared theta equal to cos squared theta plus sin squared theta​

Answer 1

Given:-

• $\sf {sec}^{2} (\theta) - {tan}^{2} (\theta) = {cos}^{2} (\theta) + {sin}^{2} (\theta)$

We have to prove this.

Answer:-

❖ We know that,

$\sf {sin}^{2} (\theta) + {cos}^{2} (\theta) = 1 \quad \quad \dots(1)$

❖ We also know that,

$\sf \: 1 + {tan}^{2} (\theta) = {sec}^{2}(\theta)$

$\sf \: \longrightarrow \: {sec}^{2}(\theta) - {tan}^{2} (\theta) = 1 \quad \quad \dots(2)$

❖ We can write that,

$1 = 1$

And since the RHS of both equation (1) and equation (2) are same (that is 1), we can equate them.

$\sf \longrightarrow \underline{ \underline{ \sf{ \green{ {sin}^{2} (\theta) + {cos}^{2} (\theta) = {sec}^{2} (\theta) - {tan}^{2}(\theta) }}}}$

Hence proved!

Other identities:-

• 1 + cot²(θ) = cosec²(θ)
• cosec(θ) = 1/sin(θ)
• sec(θ) = 1/cos(θ)
• cot(θ) = 1/tan(θ)