Question

$\large\mathbf{HEY\: QUESTION ★}$



SOLVE THE QUEST PLOXX :(​

Answer 1

$\bold{\large{\boxed{\red{ANSWER}}}}$

${( \frac{tan \: 20}{cosec \: 70} )}^{2} + {( \frac{cot \: 20}{sec \: 70}) }^{2} + 2 \: tan75 . \: tan45. \: tan15$

Using Co- function Identities :-

$\rightarrow {( \frac{cot \: (90 - 20)}{cosec \: 70}) }^{2} + {( \frac{tan\: (90 - 20)}{sec \: 70}) }^{2} + 2cot \: (90 - 75). \: (1). \: tan15$

$\rightarrow ({ \frac{cot \: 70}{cosec \: 70} )}^{2} + ({ \frac{tan \: 70}{sec \: 70} )}^{2} + 2cot \: 15. \: tan15$

Using Reciprocal Identities :-

$\rightarrow {( \frac{ \frac{cos \: 70}{sin \: 70} }{ \frac{1}{sin \: 70} }) }^{2} + {( \frac{ \frac{sin \: 70}{cos \: 70} }{ \frac{1}{cos \: 70} } }^{2} ) + 2 \frac{1}{tan \: 15} .tan \: 15$

$\rightarrow \: {(cos \: 70)}^{2} + {(sin \: 70)}^{2} + 2$

$\rightarrow \: {cos}^{2} 70 + {sin}^{2} 70 + 2$

Using Pythagorean Identities :-

$\rightarrow1 + 2$

$\rightarrow3$

The same answer is also in the attachement above !!!

$\bold {\large{\blue{\tt{I \: hope \: it \: helps \: u \: dear \: friend \: !!!}}}}$

Answer 2

( csc70∘tan20° ) 2 +( sec70 ∘cot20°)2 + 2tan15° tan45 ∘

tan75°=(

csc70∘tan(90° −70°)2 +(sec70°cot(90∘

−70° )2 +2tan(90 ∘

−75∘ )tan45 ∘ tan75 ∘=( csc70 ∘cot70° )2

+(sec70 ∘tan70 ∘ ) 2 +2cot75 ∘ tan45 ∘ tan75 ∘=(cot70 ∘sin70 ∘ ) 2 +(tan70 ∘ cos70∘ ) 2 +2×1×1 since tanθcotθ=1=( sin70 °cos70 °×sin70∘) 2 +(cos70 °sin70 °×cos70 ° )2+2=cos2 70° +sin 2 70 ° 2+2=1+2=3