Question

sin 35^ cos 55^+cos 35^ sin 55^/ cosec2 10^- tan2 80^

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Answer 1

Answer:

1 is the answer

Step-by-step explanation:

Answer 2

Step-by-step explanation:

Trigonometric Problem

Step-by-step explanation:

$\therefore \frac{sin35^{\circ}\:cos55^{\circ}+cos35^{\circ}\:sin55^{\circ}}{cosec^{2}10^{\circ}-tan^{2}80^{\circ}}$

∴ cosec 2 10∘ −tan 2 80 ∘ sin35 ∘ cos55 ∘ +cos35 ∘sin55 ∘

$=\frac{sin(35^{\circ}+55^{\circ})}{cosec^{2}10^{\circ}-tan^{2}(90^{\circ}-10^{\circ})}$= cosec 2 10 ∘ −tan 2 (90 ∘ −10 )sin(35 ∘ +55 ∘ )

$=\frac{sin90^{\circ}}{cosec^{2}10^{\circ}-cot^{2}10^{\circ}}$

cosec 2 10 ∘ −cot 2 10 ∘

sin90∘

$=\frac{1}{1}=$

11

=1=1

$\Rightarrow \boxed{\frac{sin35^{\circ}\:cos55^{\circ}+cos35^{\circ}\:sin55^{\circ}}{cosec^{2}10^{\circ}-tan^{2}80^{\circ}}=1}⇒ </p><p>cosec$

2 10 ∘ −tan 2 80 ∘

cos55 ∘ +cos35

∘ sin55∘

=1

Rules used in solving the problem:

1.$\:sinA\:cosB+cosA\:sinB=sin(A+B)1.sinAcosB+cosAsinB=sin(A+B)$

2.$\:tan(90^{\circ}-A)=cotA2.tan(90 ∘ −A)=cotA$

3.$\:cosec^{2}A-cot^{2}A=13.cosec 2 A−cot 2 A=1$

Note:

The most important thing in solving trigonometric problems is to use its identities or formulas appropriately.

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