KnowledgeBoat Logo

Mathematics

Evaluate :

sec 42°cosec 48°+3 tan 50°cot 40°−2 cos 43°sin 47°\dfrac{\text{sec } 42°}{\text{cosec } 48°} + \dfrac{3 \text{ tan } 50°}{\text{cot } 40°} - \dfrac{2 \text{ cos } 43°}{\text{sin } 47°}

Trigonometrical Ratios

1 Like

Answer

=sec 42°cosec 48°+3 tan 50°cot 40°−2 cos 43°sin 47°=sec (90°−48°)cosec 48°+3 tan (90°−40°)cot 40°−2 cos (90°−47°)sin 47°=cosec 48°cosec 48°+3 cot 40°cot 40°−2 sin 47°sin 47°=1+3−2=4−2=2= \dfrac{\text{sec } 42°}{\text{cosec } 48°} + \dfrac{3 \text{ tan } 50°}{\text{cot } 40°} - \dfrac{2 \text{ cos } 43°}{\text{sin } 47°}\\[1em] = \dfrac{\text{sec } (90° - 48°)}{\text{cosec } 48°} + \dfrac{3 \text{ tan } (90° - 40°)}{\text{cot } 40°} - \dfrac{2 \text{ cos } (90° - 47°)}{\text{sin } 47°}\\[1em] = \dfrac{\text{cosec } 48°}{\text{cosec } 48°} + \dfrac{3 \text{ cot } 40°}{\text{cot } 40°} - \dfrac{2 \text{ sin } 47°}{\text{sin } 47°}\\[1em] = 1 + 3 - 2\\[1em] = 4 - 2\\[1em] = 2

Hence, the value of sec 42°cosec 48°+3 tan 50°cot 40°−2 cos 43°sin 47°=2\dfrac{\text{sec } 42°}{\text{cosec } 48°} + \dfrac{3 \text{ tan } 50°}{\text{cot } 40°} - \dfrac{2 \text{ cos } 43°}{\text{sin } 47°} = 2.

Answered By

1 Like

Related Questions