Question 2 Prove that- - 2-cosec2-tan-

LHS=√(sec^2 θ+cosec^2 θ) =√(1/cos^2 θ+1/sin^2 θ) [ ∵ sec θ=1/cos θ and cosec θ=1/sin θ ] =√(sin^2 θ+cos^2 θ/sin^2 θ.cos^2 θ)=√(1/sin^2 θ. cos^2 θ) [∵ sin^ ...

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Byju's Answer

Standard X

Mathematics

Trigonometric Identities

Question 2 Pr...

Question

Question 2
Prove that: $\sqrt{s e c^{2} θ + c o s e c^{2} θ} = t a n θ + c o t θ$

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Solution

$L H S = \sqrt{s e c^{2} θ + c o s e c^{2} θ}$

$= \sqrt{\frac{1}{c o s^{2} θ} + \frac{1}{s i n^{2} θ}} [∵ s e c θ = \frac{1}{c o s θ} a n d c o s e c θ = \frac{1}{s i n θ}]$

$= \sqrt{\frac{s i n^{2} θ + c o s^{2} θ}{s i n^{2} θ . c o s^{2} θ}} = \sqrt{\frac{1}{s i n^{2} θ . c o s^{2} θ}}$ $[∵ s i n^{2} θ + c o s^{2} θ = 1]$

$= \sqrt{\frac{s i n^{2} θ + c o s^{2} θ}{s i n^{2} θ . c o s^{2} θ}} = \sqrt{\frac{1}{s i n^{2} θ . c o s^{2} θ}}$ $[∵ 1 = s i n^{2} θ + c o s^{2} θ]$

$= \frac{s i n^{2} θ}{s i n θ . c o s θ} + \frac{c o s^{2} θ}{s i n θ . c o s θ}$

$= \frac{s i n θ}{c o s θ} + \frac{c o s θ}{s i n θ}$ $[∵ t a n θ = \frac{s i n θ}{c o s θ} a n d c o t θ = \frac{c o s θ}{s i n θ}]$

$= t a n θ + c o t θ = R H S$

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Similar questions

Q. Prove the following :
(viii)

\sqrt{s e c^{2} θ + \cos e c^{2} θ} = \tan θ + c o t θ

Q. Prove that

\sqrt{{sec}^{2} θ + c o s e c^{2} θ} = tan θ + cot θ

Q. If tan θ =

\frac{1}{\sqrt{7}}

then prove that

(\frac{{cosec}^{2} θ + \sec^{2} θ}{{cosec}^{2} θ - \sec^{2} θ}) = \frac{4}{3} .

Q. Prove the following identity.

\sqrt{{sec}^{2} θ + c o s e c^{2} θ} = tan θ + cot θ

If $θ$ is acute angle, and $t a n θ$ = $\frac{1}{\sqrt{7}}$ , then $\frac{c o s e c^{2} θ - s e c^{2} θ}{c o s e c^{2} θ + s e c^{2} θ} =$

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Question 2 Prove that: √sec2 θ+cosec2 θ=tan θ+cot θ

Question 2
Prove that: $\sqrt{s e c^{2} θ + c o s e c^{2} θ} = t a n θ + c o t θ$