The minimum value of sin2- -cos2- -2- -cosec2- -tan2- -2-

sin ^ 2 θ + cos ^ 2 θ + sec ^ 2 θ + cosec ^ 2 θ + tan ^ 2 θ + cot ^ 2 θ #160; 6 ≥√( sin ^ 2 θ cos ^ 2 θ tan ^ 2 θ #160; / sin ^ 2 θ cos ^ 2 θ tan ^ 2 ...

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Byju's Answer

Standard XII

Mathematics

Sign of Trigonometric Ratios in Different Quadrants

The minimum v...

Question

The minimum value of ${sin}^{2} θ + {cos}^{2} θ + {sec}^{2} θ + c o s e c^{2} θ + {tan}^{2} θ + {cot}^{2} θ$ .

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Solution

$\frac{{s i n}^{2} θ + {c o s}^{2} θ + {s e c}^{2} θ + {c o s e c}^{2} θ + {t a n}^{2} θ + {c o t}^{2} θ}{6} \geq \sqrt[6]{\frac{{s i n}^{2} θ {c o s}^{2} θ {t a n}^{2} θ}{{s i n}^{2} θ {c o s}^{2} θ {t a n}^{2} θ}} {s i n}^{2} θ + {c o s}^{2} θ + {s e c}^{2} θ + {c o s e c}^{2} θ + {t a n}^{2} θ + {c o t}^{2} θ \geq 6 \sqrt[6]{1}$
minimum value 6

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Similar questions

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. Prove that:

(i)

s e c^{2} θ + c o s e c^{2} θ = s e c^{2} θ c o s e c^{2} θ

(ii)

t a n^{2} θ - s i n^{2} θ = t a n^{2} θ s i n^{2} θ

(iii)

t a n^{2} θ + c o t^{2} θ + 2 = s e c^{2} θ c o s e c^{2} θ

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

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

Q. Find the value

\sqrt{1 + t a n^{2} θ} \sqrt{1 + c o t^{2} θ} \sqrt{1 - c o s^{2} θ} \sqrt{1 - s i n^{2} θ} =

Q. Derive the identify

1 + {tan}^{2} θ = {sec}^{2} θ

from the identify

{sin}^{2} θ + {cos}^{2} θ = 1

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The minimum value of sin2θ+cos2θ+sec2θ+cosec2θ+tan2θ+cot2θ.

sin2θ+cos2θ+sec2θ+cosec2θ+tan2θ+cot2θ6≥6√sin2θcos2θtan2θsin2θcos2θtan2θsin2θ+cos2θ+sec2θ+cosec2θ+tan2θ+cot2θ≥66√1minimum value 6

The minimum value of ${sin}^{2} θ + {cos}^{2} θ + {sec}^{2} θ + c o s e c^{2} θ + {tan}^{2} θ + {cot}^{2} θ$ .