Question

26. Find the value of (i) sin 38˚36′ + tan 12˚12′ (ii) tan 60˚ 25′ – cos 49˚20′.​

Answer 1

Answer:

(i) sin26°cos64°

sin26°cos64°=sin90°-64°cos64°            =cos64°cos64°                 ∵ sin90°-θ=cosθ            =1Hence, sin26°cos64°=1

(ii) sec11∘cosec79∘      =sec(90∘−79∘)cosec79∘        =cosec79∘cosec79∘         [∵sec 90-θ = cosec θ]=1      

(iii) tan65°cot25°

tan65°cot25°=tan90°-25°cot25°            =cot25°cot25°                 ∵ tan90°-θ=cotθ            =1Hence, tan65°cot25°=1

(iv) cos37°sin53°

cos37°sin53°=cos90°-53°sin53°            =sin53°sin53°                 ∵ cos90°-θ=sinθ            =1Hence, cos37°sin53°=1

(v)cosec42∘sec48∘    =cosec(90∘−48∘)sec48∘        =sec48∘sec48∘        [∵sec 90-θ = cosec θ]    =1

(vi) cot34°tan56°

cot34°tan56°=cot90°-56°tan56°            =tan56°tan56°                 ∵ cot90°-θ=tanθ            =1Hence, cot34°tan56°=1

Step-by-step explanation:

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