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Q2. In Fig. find tan P – cot R.



Answer :


By using the Pythagoras theorem, we have

\( PR^2 \ = \ PQ^2 \ + \ QR^2 \)

=> \( 13^2 \ = \ 12^2 \ + \ QR^2 \)

=> \( QR^2 \ = \ 13^2 \ - \ 12^2 \ = \ 169 \ - \ 144 \ = \ 25 \)

=> \( QR \ = \ \sqrt{25} \ = \ 5 \)

∴ \( tanP \ = \ \frac{P}{B} \ = \ \frac{QR}{PQ} \ = \ \frac{5}{12} \)
\( cotR \ = \ \frac{B}{P} \ = \ \frac{QR}{PQ} \ = \ \frac{5}{12} \)

Hence, \( tan P \ – \ cot R \ = \ \frac{5}{12} \ - \ \frac{5}{12} \ = \ 0 \)