I 1- cot2-1-cosec -cosec - ii 1- tan2-1-sec -sec -

(i) 1 + cot^2 θ/1+ cosec θ = cosec θ LHS = 1 + cot^2 θ/1+ cosec θ = 1 + cosec^2 θ 1/cosec θ+1 = 1 + (cosec θ+1)(cosec θ 1)/cosec θ+1 = 1 + (cosec θ 1) = ...

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Byju's Answer

Standard X

Mathematics

Trigonometric Identities

i 1+ cot2θ/1+...

Question

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

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Solution

(i) $1 + \frac{c o t^{2} θ}{1 + c o s e c θ} = c o s e c θ L H S = 1 + \frac{c o t^{2} θ}{1 + c o s e c θ} = 1 + \frac{c o s e c^{2} θ - 1}{c o s e c θ + 1} = 1 + \frac{(c o s e c θ + 1) (c o s e c θ - 1)}{c o s e c θ + 1} = 1 + (c o s e c θ - 1) = c o s e c θ = R H S$

(ii) $1 + \frac{t a n^{2} θ}{1 + s e c θ} = s e c θ L H S = 1 + \frac{t a n^{2} θ}{1 + s e c θ} = 1 + \frac{s e c^{2} θ - 1}{s e c θ + 1} = 1 + \frac{(s e c θ + 1) (s e c θ - 1)}{s e c θ + 1} = 1 + (s e c θ - 1) = s e c θ = R H S$

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Similar questions

Q. Find

θ -

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

\sqrt{\frac{1 + \sin θ}{1 - \sin θ}}

is equal to

(a) sec θ + tan θ
(b) sec θ − tan θ
(c) sec² θ + tan² θ
(d) sec² θ − tan² θ

Q. More than One Answer Type
एक से अधिक उत्तर प्रकार के प्रश्न

If

x = (\frac{{sin}^{2} θ {sec}^{2} θ + 1}{{sec}^{2} θ}) + (\frac{{cot}^{2} θ}{1 + c o s e c θ}),

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

and

z = \frac{1 - {sin}^{2} θ + sin θ cos θ}{sin θ - {sin}^{3} θ},

where 0° < θ < 90°, then the correct option(s) is/are

यदि

x = (\frac{{sin}^{2} θ {sec}^{2} θ + 1}{{sec}^{2} θ}) + (\frac{{cot}^{2} θ}{1 + c o s e c θ}),

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

तथा

z = \frac{1 - {sin}^{2} θ + sin θ cos θ}{sin θ - {sin}^{3} θ},

जहाँ 0° < θ < 90°, तब सही विकल्प है/हैं

Q. Is LHS=RHS?

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

Say true or false

Q. Given that

cos

θ = \frac{1}{2}

, then the value of

\frac{2 sec θ}{1 + {tan}^{2} θ}

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(i) 1+cot2θ(1+cosecθ)=cosecθ (ii) 1+tan2θ(1+secθ)=secθ

(i) 1+cot2 θ1+cosec θ=cosec θLHS=1+cot2 θ1+cosec θ=1+cosec2 θ−1cosec θ+1=1+(cosec θ+1)(cosec θ−1)cosec θ+1=1+(cosec θ−1)=cosec θ=RHS (ii)1+tan2 θ1+sec θ=sec θLHS=1+tan2 θ1+sec θ=1+sec2 θ−1sec θ+1=1+(sec θ+1)(sec θ−1)sec θ+1=1+(sec θ−1)=sec θ=RHS

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