Question

(sec theta + cos theta) (sec theta- cos theta)= tan^2theta+ sin^2theta​

Answer 1

Step-by-step explanation:

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Answer 2

Topic:-

Trigonometry

Question:-

$prove \: that \\ \\ (sec \theta + cos \theta)(sec \theta - cos \theta) = {tan}^{2} \theta + {sin}^{2} \theta$

Solution:-

$take \: lhs \: \\ \\ (sec \theta + cos \theta)(sec \theta - cos \theta) \\ \\ we \: know \: that \: (a + b)(a - b) = {a}^{2} - {b}^{2} \\ \\ = {sec}^{2} \theta - {cos}^{2} \theta \\ \\ = (1 + {tan}^{2} \theta) - (1 - {sin}^{2} \theta) \\ \\ = 1 + {tan}^{2} \theta - 1 + {sin}^{2} \theta \\ \\ = \cancel{1} + {tan}^{2} \theta \: \cancel{ - 1} + {sin}^{2} \theta \\ \\ = {tan}^{2} \theta + {sin}^{2} \theta \\ \\ = rhs \\ \\ hence \: proved$

More Information:-

Trigon metric Identities

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

csc²θ - cot²θ = 1

Trigometric relations

sinθ = 1/cscθ

cosθ = 1 /secθ

tanθ = 1/cotθ

tanθ = sinθ/cosθ

cotθ = cosθ/sinθ

Trigonmetric ratios

sinθ = opp/hyp

cosθ = adj/hyp

tanθ = opp/adj

cotθ = adj/opp

cscθ = hyp/opp

secθ = hyp/adj