Question

Sec^2 theta =3 find value of {tan^2-sec^2theta÷tan^2theta +cos^2theta}

Answer 1

ANSWER:

Given that

sec²θ = 3

secθ = √3 = √3/1

Hence,

Hypotenuse = √3

Base = 1

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Using, Pythagoras theorem

(Hypotenuse)² = (Base)² + (Height)²

(√3)² = 1² + (h)²

3 - 1 = h²

h = √2

TO FIND:

tan²θ - sec²θ/tan²θ + cos²θ

Now,

tanθ = opposite/adjacent = √2/1 = √2

secθ = √3 (.°. Given)

cosθ = 1/√3

Because secθ & cosθ are reciprocal to each other

(√2)² - (√3)²/(√2)² + (1/√3)²

= 2 - 3/2 + 1/3

= - 1/7/3

= - 3/7

.°.Answer is - 3/7