দেখাই যে : \(cosec^2 48°– \tan^2 42°=1\)
\(cosec ^2 48°-tan^2 42° \)
\(=cosec^2 (90°-42°)-tan^2 42° \)
\(=sec^2 42° -tan^2 42° [∵cosec (90°-θ)=secθ] \)
\(=1\) (প্রমাণিত) \( [∵sec^2θ-tan^2θ=1 ]\)
Similar Questions
1.
\(cosec^2 48°– \tan^2 42°=1\)