Question 2Prove that- -sec2 - - cosec2 - - tan - - cot -

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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Standard XII

Mathematics

Trigonometric Ratios Using Right Angled Triangle

Question 2Pro...

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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Q. Question 15
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t a n^{4} θ + t a n^{2} θ = s e c^{4} 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: $\sqrt{s e c^{2} θ + c o s e c^{2} θ} = t a n θ + c o t θ$