Question

१) गोपाल प्रसाद के अनुसार दगा देने वाला --

a) तांगेवाला

b)रामस्वरुप

c)गोपाल प्रसाद

fast

it should be correct ​​

Answer 1

$\huge\bold{Question :}$

If tan θ = ¹/√7 then , show that $\sf \dfrac{csc^2\theta-sec^2\theta}{csc^2\theta+sec^2\theta}=\dfrac{3}{4}$

$\huge\bold{Solution :}$

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$\sf tan\ \theta=\dfrac{1}{\sqrt{7}}$

$:\to \sf tan^2\theta=\dfrac{1}{(\sqrt{7})^2}$

$:\to \sf \textsf{\textbf{\pink{tan ^\text{2} \boldsymbol \theta\ =\ \dfrac{\text{1}}{\text{7}} }}}\ \; \bigstar$

$\sf \dfrac{1}{cot\ \theta}=\dfrac{1}{\sqrt{7}}$

$:\to \sf cot\ \theta=\sqrt{7}$

$:\to \sf cot^2\theta=(\sqrt{7})^2$

$:\to \sf \textsf{\textbf{\green{cot ^\text{2}\ \boldsymbol \theta \ =\ 7}}}\ \; \bigstar$

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LHS

$:\to \bf \blue{\dfrac{csc^2\theta-sec^2\theta}{csc^2\theta+sec^2\theta}}$

From Trigonometric identities ,

csc²θ = 1 + cot²θ

sec²θ = 1 + tan²θ

$:\to \sf \dfrac{(1+cot^2\theta)-(1+tan^2\theta)}{(1+cot^2\theta)+(1+tan^2\theta)}$

tan²θ = ¹/₇

cot²θ = 7

$:\to \sf \dfrac{(1+7)-(1+\frac{1}{7})}{(1+7)+(1+\frac{1}{7})}$

$:\to\ \sf \dfrac{8-\frac{8}{7}}{8+\frac{8}{7}}$

$:\to\ \sf \dfrac{48}{64}$

$:\to\ \textsf{\textbf{\orange{ \dfrac{\text{3}}{\text{4}} }}}\ \; \bigstar$

Answer 2

Answer:

b) रामस्वरूप है।

Hope this helps you brother