Question

✌❤ plz Solve it......✌❤​

Answer 1

To prove --->

${sec}^{2} \alpha \: - \: {cos}^{2} \beta \: = {sin}^{2} \beta \: + \: {tan}^{2} \alpha$

Concept used ---->

1)

$1 \: + \: {tan}^{2} \gamma \: = \: {sec}^{2} \gamma$

2)

$\ {sin}^{2} \gamma \: + \: {cos}^{2} \gamma \: = \: 1 \\ = > \: {cos}^{2} \gamma \: = \: 1 \: - \: {sin}^{2} \gamma$

Proof ---> LHS

$= \: {sec}^{2} \alpha \: - \: {cos}^{2} \beta \\ putting \: {sec}^{2} \alpha \: = \: 1 \: + {tan}^{2} \alpha \: and \: {cos}^{2} \beta \: = \: 1 \: - \: {sin}^{2} \beta \: we \: get \\ = \: 1 \: + \: {tan}^{2} \alpha \: - \: ( \: 1 \: - \: {sin}^{2} \beta \: ) \\ = \: 1 \: + {tan}^{2} \alpha \: - \: 1 \: + {sin}^{2} \beta \\ + 1 \: and \: - 1 cancel \: out \: each \: other \\ = {sin}^{2} \beta \: + \: {tan}^{2} \alpha$

= RHS

Additional information ---->

1)

$1 \: + \: {cot}^{2} \gamma \: = \: {cosec}^{2} \gamma$

Answer 2

SOLUTION

To Prove

$\sf \sec {}^{2} x - {cos}^{2} y = {sin}^{2} y + {tan}^{2} x$

Identities Used

sin²∅ + cos²∅ = 1

• cos²∅ = 1 - sin²∅

sec²∅ - tan²∅ = 1

• sec²∅ = 1 + tan²∅

From the above relations,

$\longrightarrow \: \sf \: (1 + {tan}^{2} x) - (1 - {sin}^{2} y) \\ \\ \longrightarrow \: \sf \: \cancel{1 - 1} \: + tan {}^{2} x - ( - {sin}^{2} y) \\ \\ \longrightarrow \: \sf \: {tan}^{2} x + {sin}^{2} y \longrightarrow RHS$

Henceforth, Proved