Find the value of -1-tan2 -1-2 -1-cos2 -1-sin2 -

Note that 1+tan^2 θ = sec^2 θ, sin^2 θ+ cos^2 θ=1, and 1+cot^2 θ=cosec^2 θ ∴√(1+tan^2θ)×√(1+cot^2θ)×√(1 cos^2θ)×√(1 sin^2θ) =√(sec^2θ)×√(cosec^2θ)×√(sin^2θ)×√( ...

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Byju's Answer

Standard X

Mathematics

Trigonometric Ratio(sec,cosec,cot)

Find the valu...

Question

Find the value of $\sqrt{1 + {tan}^{2} θ} \sqrt{1 + {cot}^{2} θ} \sqrt{1 - {cos}^{2} θ} \sqrt{1 - {sin}^{2} θ}$ .

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Solution

Note that $1 + t a n^{2} θ = s e c^{2} θ$ , $s i n^{2} θ + c o s^{2} θ = 1$ , $a n d 1 + c o t^{2} θ = c o s e c^{2} θ$
$∴ \sqrt{1 + t a n^{2} θ} \times \sqrt{1 + c o t^{2} θ} \times \sqrt{1 - c o s^{2} θ} \times \sqrt{1 - s i n^{2} θ} = \sqrt{s e c^{2} θ} \times \sqrt{c o s e c^{2} θ} \times \sqrt{s i n^{2} θ} \times \sqrt{c o s^{2} θ} = s e c θ \times c o s e c θ \times s i n θ \times c o s θ = (s e c θ \times c o s θ) \times (c o s e c θ \times s i n θ) = 1 \times 1 = 1$

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Similar questions

\frac{(1 - \tan^{2} θ)}{(1 + \tan^{2} θ)} = ?

(a) (sin²θ − cos²θ)
(b) (cos²θ − sin²θ)
(c) (cot²θ − tan²θ)
(d) (tan²θ − cot²θ)

Q. Prove the following identity

\frac{{tan}^{2} θ}{{tan}^{2} θ - 1} + \frac{c o s e c^{2} θ}{{sec}^{2} θ - c o s e c^{2} θ} = \frac{1}{{sin}^{2} θ - {cos}^{2} θ}

Q. Derive the identify

1 + {tan}^{2} θ = {sec}^{2} θ

from the identify

{sin}^{2} θ + {cos}^{2} θ = 1

Q. The minimum value of

{sin}^{2} θ + {cos}^{2} θ + {sec}^{2} θ + c o s e c^{2} θ + {tan}^{2} θ + {cot}^{2} θ

Q. If

s i n θ = \frac{12}{13},

find the value of

\frac{s i n^{2} θ - c o s^{2} θ}{2 s i n θ c o s θ} \times \frac{1}{t a n^{2} θ}

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Find the value of √1+tan2θ√1+cot2θ√1−cos2θ√1−sin2θ.

Note that 1+tan2θ=sec2θ, sin2θ+cos2θ=1, and 1+cot2θ=cosec2θ ∴√1+tan2θ×√1+cot2θ×√1−cos2θ×√1−sin2θ=√sec2θ×√cosec2θ×√sin2θ×√cos2θ=secθ×cosecθ×sinθ×cosθ=(secθ×cosθ)×(cosecθ×sinθ)=1×1=1

Find the value of $\sqrt{1 + {tan}^{2} θ} \sqrt{1 + {cot}^{2} θ} \sqrt{1 - {cos}^{2} θ} \sqrt{1 - {sin}^{2} θ}$ .