Question

active member in case of Hindu Undivided family business Comment
22 Harish took a fire insurance policy of rupees 20 lakh for his factory at the annual premium of rupees 24,000. In
order to avoid premium more than this amount, he did not disclose that highly explosive chemicals are being
manufactured in bis factory. Due to a fire, his factory gets severely damaged. The insurance company refused to
make the payment for claim as it became aware
about the bighly explosive chemicals. Ls Harish entitled to receive the claim? Explain the principle
of insurance violated by Harish
3h​

Answer 1

Answer:

$Question:</p><p></p><p>If tan θ = ¹/√7 then , show that \sf \dfrac{csc^2\theta-sec^2\theta}{csc^2\theta+sec^2\theta}=\dfrac{3}{4}csc2θ+sec2θcsc2θ−sec2θ=43</p><p></p><p>\huge\bold{Solution :}Solution:</p><p></p><p>★══════════════════════★</p><p></p><p>\sf tan\ \theta=\dfrac{1}{\sqrt{7}}tan θ=71</p><p></p><p>:\to \sf tan^2\theta=\dfrac{1}{(\sqrt{7})^2}:→tan2θ=(7)21</p><p></p><p>:\to \sf \textsf{\textbf{\pink{tan ^\text{2} \boldsymbol \theta\ =\ \dfrac{\text{1}}{\text{7}} }}}\ \; \bigstar:→tan2θ  = 71 ★</p><p></p><p>\sf \dfrac{1}{cot\ \theta}=\dfrac{1}{\sqrt{7}}cot θ1=71</p><p></p><p>:\to \sf cot\ \theta=\sqrt{7}:→cot θ=7</p><p></p><p>:\to \sf cot^2\theta=(\sqrt{7})^2:→cot2θ=(7)2</p><p></p><p>:\to \sf \textsf{\textbf{\green{cot ^\text{2}\ \boldsymbol \theta \ =\ 7}}}\ \; \bigstar:→cot2 θ = 7 ★</p><p></p><p>★══════════════════════★</p><p></p><p>LHS</p><p></p><p>:\to \bf \blue{\dfrac{csc^2\theta-sec^2\theta}{csc^2\theta+sec^2\theta}}:→csc2θ+sec2θcsc2θ−sec2θ</p><p></p><p>From Trigonometric identities ,</p><p></p><p>csc²θ = 1 + cot²θ</p><p></p><p>sec²θ = 1 + tan²θ</p><p></p><p>:\to \sf \dfrac{(1+cot^2\theta)-(1+tan^2\theta)}{(1+cot^2\theta)+(1+tan^2\theta)}:→(1+cot2θ)+(1+tan2θ)(1+cot2θ)−(1+tan2θ)</p><p></p><p>tan²θ = ¹/₇</p><p></p><p>cot²θ = 7</p><p></p><p>:\to \sf \dfrac{(1+7)-(1+\frac{1}{7})}{(1+7)+(1+\frac{1}{7})}:→(1+7)+(1+71)(1+7)−(1+71)</p><p></p><p>:\to\ \sf \dfrac{8-\frac{8}{7}}{8+\frac{8}{7}}:→ 8+788−78</p><p></p><p>:\to\ \sf \dfrac{48}{64}:→ 6448</p><p></p><p>:\to\ \textsf{\textbf{\orange{ \dfrac{\text{3}}{\text{4}} }}}\ \; \bigstar:→ 43 ★</p><p></p><p>$