Question

If cos theta + sec theta equals to 3 then find the value of cos square theta plus sec square theta

Answer 1

Question :-

If $\sf{cos\theta+sec\theta=3}$, then find the value of $\sf{cos^2\theta+sec^2\theta}$.

Solution :-

$\sf{cos\theta+sec\theta=3}$

$\to\sf{(cos\theta+sec\theta)^2=3^2\:[Squaring\:both\: sides]}$

$\to\sf{cos^2\theta+sec^2\theta+2.\:cos\theta.\:sec\theta=9}$

$\to\sf{cos^2\theta+sec^2\theta+2.\:cos\theta\times\dfrac{1}{cos\theta}=9}$

$\to\sf{cos^2\theta+sec^2\theta+2\times\:1=9}$

$\to\sf{cos^2\theta+sec^2\theta+2=9}$

$\to\sf{cos^2\theta+sec^2\theta=9-2}$

$\to\sf{cos^2\theta+sec^2\theta=7\:[Answer]}$

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Some formulas :-

★ sin²A + cos²A = 1

★ 1 + tan²A = sec²A

★ 1 + cot²A = cosec²A

★ cos²A - sin²A = cos2A

★ sin(A+B) = sinAcosB + cosAsinB

★ sin(A-B) = sinAcosB - cosAsinB

★ cos(A+B) = cosAcosB - sinAsinB

★ cos(A-B) = cosAcosB + sinAsinB

Answer 2

Given :

Cos(Φ) + Sec(Φ)= 3

To find :

Cos(Φ)² + Sec(Φ)²

Solution :

Squaring on both sides , we get

{Cos(Φ) + Sec(Φ)}² = (3)²

Cos(Φ)² + Sec(Φ)² + 2Cos(Φ)Sec(Φ) = 9

Cos(Φ)² + Sec(Φ)² + 2 × 1 = 9

Cos(Φ)² + Sec(Φ)² = 9 - 2

Cos(Φ)² + Sec(Φ)² = 7

$\therefore$ The required value is 7

Remember :

$\mapsto$ Cos(Φ) = 1/Sec(Φ)