If tan -1- a 2- prove that -tan 3-cosec-2- a 23-2-

Sec θ +tan^3 θ cosec θ =sec θ ( (sec θ+tan^3 θ cosec θ)/sec θ ) =sec θ((1+tan^3 θcos θ/sin θ) =sec θ(1+tan^3 θ× cot θ) = √(1+tan^2θ)(1+tan^2 θ) =(1+tan^2)^ ...

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Standard X

Mathematics

Trigonometric Identities

If tan θ=1- a...

Question

If $t a n^{2} θ = 1 - a^{2}$ , prove that $s e c θ + (t a n^{3} θ \times c o s e c θ) = (2 - a^{2})^{\frac{3}{2}}$ .

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Solution

$s e c θ + t a n^{3} θ c o s e c θ = s e c θ (\frac{(s e c θ + t a n^{3} θ c o s e c θ)}{s e c θ})$

$= s e c θ ((1 + t a n^{3} θ \frac{c o s θ}{s i n θ}) = s e c θ (1 + t a n^{3} θ \times c o t θ) = \sqrt{1 + t a n^{2} θ} (1 + t a n^{2} θ) = (1 + t a n^{2})^{\frac{3}{2}} = (2 - a^{2})^{\frac{3}{2}}$

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If tan2 θ=1−a2, prove that sec θ+(tan3 θ×cosec θ)=(2−a2)32.

sec θ+tan3 θ cosec θ=sec θ((sec θ+tan3 θ cosec θ)sec θ) =sec θ((1+tan3 θcos θsin θ)=sec θ(1+tan3 θ×cot θ)=√1+tan2θ(1+tan2θ)=(1+tan2)32=(2−a2)32

If $t a n^{2} θ = 1 - a^{2}$ , prove that $s e c θ + (t a n^{3} θ \times c o s e c θ) = (2 - a^{2})^{\frac{3}{2}}$ .