Question

please give me the solution fast​

Answer 1

✪AnSwEr

${\tt{\red{\underline{\large{Given}}}}}$

• (sin⁴∅-Cos⁴∅+1)Cosec²∅

${\sf{\green{\underline{\large{To\:find}}}}}$

• Value

${\sf{\pink{\underline{\Large{Explanation}}}}}$

• We know that
• a²-b²=(a+b)(a-b)
• sin²∅+cos²∅=1

Solution

$\rule{200}4$

(sin⁴∅-Cos⁴∅+1)Cosec²∅

=>{(sin²∅-Cos²∅+1)(sin²∅+Cos²∅)+1}Cosec²∅

=>{(sin²∅-Cos²∅)(1)+1}Cosec²∅

=>{sin²∅-Cos²∅)+1}Cosec²∅

=>{sin²∅-Cos²∅+1}Cosec²∅

=>{sin²∅-Cos²∅+sin²∅+Cos²∅}Cosec²∅

=>{sin²∅+sin²∅}Cosec²∅

=>{2sin²∅}Cosec²∅

=>2Sin²∅ × 1/sin²∅

=>2

$\rule{200}4$

Option c is correct

Answer 2

QuestioN :

$\tt Solve : \: \: ({sin}^{4} \theta - {cos}^{4} \theta + 1){cosec}^{2} \theta$

• ( A ) 1.
• ( B ) - 2.
• ( C ) 2.
• ( D ) 0.

AnsWer :

( C ) 2.

Formula :

• Sin² A + Cos² A = 1.
• Cosec² A = 1 / Cos² A.

SolutioN :

We have,

$\tt \longmapsto ({sin}^{4} \theta - {cos}^{4} \theta + 1){cosec}^{2} \theta$

We can also Write as,

$\tt \mapsto {sin}^{2} \theta + {cos}^{2} \theta = 1.$

$\tt \mapsto 1 - {sin}^{2} \theta = {cos}^{2} \theta .$

★ Squaring both sides, We get.

$\tt \mapsto {(1 - {sin}^{2} \theta ) }^{2} = ({ {cos}^{2} \theta )}^{2} .$

$\tt \mapsto 1 + {sin}^{4} \theta - 2 {sin}^{2} \theta = {cos}^{4} \theta.$

★ So, We can substitute Values ( Cos⁴ theta )

$\tt \longmapsto ({sin}^{4} \theta - (1 + {sin}^{4} \theta - 2 {sin}^{2} \theta) + 1){cosec}^{2} \theta.$

$\tt \longmapsto ({sin}^{4} \theta - 1 - {sin}^{4} \theta + 2 {sin}^{2} \theta+ 1){cosec}^{2} \theta.$

$\tt \longmapsto ( \cancel{{sin}^{4} \theta} \cancel{ - 1 } \cancel{- {sin}^{4} \theta } + 2 {sin}^{2} \theta \cancel{+ 1}){cosec}^{2} \theta.$

$\tt \longmapsto ( 2 {sin}^{2} \theta){cosec}^{2} \theta.$

We know,

• 1 / Sin² A = Cosec² A.

$\tt \longmapsto 2 {sin}^{2} \theta \times \dfrac{1}{{sin }^{2} \theta}$

$\tt \longmapsto 2 \cancel{ {sin}^{2} \theta} \times \dfrac{1}{ \cancel{{sin }^{2} \theta}}$

$\tt \longmapsto 2 .$

Therefore the required value is 2.