Prove that sin 49- tan 11- sec 41- tan 79-1 -2 MARKS-

Concept: 1 Mark Application: 1 Mark Consider L.H.S sin 49^∘. tan 11^∘. sec 41^∘. tan 79^∘ ⇒ sin 49^∘. sec 41^∘. tan 11^∘. tan 79^∘ = sin 49^∘. sec (90 49^ ...

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Standard XII

Mathematics

Trigonometric Ratios of Common Angles

Prove that si...

Question

Prove that $s i n 49^{\circ} . t a n 11^{\circ} . s e c 41^{\circ} . t a n 79^{\circ} = 1$ [2 MARKS]

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Solution

Concept: 1 Mark
Application: 1 Mark

Consider L.H.S

$s i n 49^{\circ} . t a n 11^{\circ} . s e c 41^{\circ} . t a n 79^{\circ}$

$\Rightarrow s i n 49^{\circ} . s e c 41^{\circ} . t a n 11^{\circ} . t a n 79^{\circ}$

$= s i n 49^{\circ} . s e c (90 - 49^{\circ}) . t a n 11^{\circ} . t a n (90 - 11^{\circ})$

$= s i n 49^{\circ} . c o s e c 49^{\circ} . t a n 11^{\circ} . c o t 11^{\circ}$

$= s i n 49^{\circ} . \frac{1}{s i n 49^{\circ}} . t a n 11^{\circ} . \frac{1}{t a n 11^{\circ}}$

$= 1$

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Prove that sin 49∘.tan 11∘.sec 41∘.tan 79∘=1 [2 MARKS]

Concept: 1 Mark Application: 1 Mark Consider L.H.S sin 49∘.tan 11∘.sec 41∘.tan 79∘ ⇒sin 49∘.sec 41∘.tan 11∘.tan 79∘ =sin 49∘.sec (90−49∘).tan 11∘.tan (90−11∘) =sin 49∘.cosec 49∘.tan 11∘.cot 11∘ =sin 49∘.1sin 49∘.tan 11∘.1tan 11∘ =1

Prove that $s i n 49^{\circ} . t a n 11^{\circ} . s e c 41^{\circ} . t a n 79^{\circ} = 1$ [2 MARKS]