Given: tan‌θ+sec‌θ=3 Formula Used: sec2θ−tan2θ=1 (sec‌θ+tan‌θ)(sec‌θ−tan‌θ)=1 Calculation: We have tan‌θ+sec‌θ=3‌‌⋯ (i) According to the formula used (sec‌θ+tan‌θ)(sec‌θ−tan‌θ)=1 ⇒3(sec‌θ−tan‌θ)=1 ⇒(sec‌θ−tan‌θ)=1∕3‌‌‌ - (ii) ‌ On adding (i) and (ii), we get 2‌sec‌θ=3+1∕3=10∕3 ⇒sec‌θ=5∕3‌‌⋯‌-(iii) ‌ On putting the value of sec‌θ in (i), we get tan‌θ+5∕3=3 ⇒tan‌θ=3−5∕3=4∕3‌‌‌ - -(iv) ‌ Now, we have to find the value of 3‌tan‌θ+9‌sec‌θ So, from (iii) and (iv), we get 3×4∕3+9×5∕3 ⇒4+15=19 ∴ The required value of 3‌tan‌θ+9‌sec‌θ is 19 .