Prove that $\sec A(1-\sin A)(\sec A+\tan A)=1$
Prove that $\sec A(1-\sin A)(\sec A+\tan A)=1$
$LHS =\sec A (1-\sin A )(\sec A +\tan A )$
$=\left(\frac{1}{\cos A }\right)(1-\sin A )\left(\frac{1}{\cos A }+\frac{\sin A }{\cos A }\right)$
$=\frac{(1-\sin A)(1+\sin A)}{\cos ^{2} A}=\frac{1-\sin ^{2} A}{\cos ^{2} A}$
$=\frac{\cos ^{2} A}{\cos ^{2} A}=1=R H S$
Similar Questions
In triangle $ABC ,$ right -angled at $B ,$ if $\tan A =\frac{1}{\sqrt{3}},$ find the value of:
$(i)$ $\sin A \cos C+\cos A \sin C$
$(ii)$ $\cos A \cos C-\sin A \sin C$
In triangle $ABC ,$ right -angled at $B ,$ if $\tan A =\frac{1}{\sqrt{3}},$ find the value of:
$(i)$ $\sin A \cos C+\cos A \sin C$
$(ii)$ $\cos A \cos C-\sin A \sin C$
State whether the following are true or false. Justify your answer.
$\sin \theta=\cos \theta$ for all values of $\theta$
State whether the following are true or false. Justify your answer.
$\sin \theta=\cos \theta$ for all values of $\theta$
In $\triangle$ $PQR,$ right-angled at $Q$ (see $Fig.$), $PQ =3 \,cm$ and $PR =6 \,cm$. Determine $\angle QPR$ and $\angle PRQ$.
In $\triangle$ $PQR,$ right-angled at $Q$ (see $Fig.$), $PQ =3 \,cm$ and $PR =6 \,cm$. Determine $\angle QPR$ and $\angle PRQ$.
Prove that
$\frac{\sin \theta-\cos \theta+1}{\sin \theta+\cos \theta-1}=\frac{1}{\sec \theta-\tan \theta},$ using the identity
$\sec ^{2} \theta=1+\tan ^{2} \theta$
Prove that
$\frac{\sin \theta-\cos \theta+1}{\sin \theta+\cos \theta-1}=\frac{1}{\sec \theta-\tan \theta},$ using the identity
$\sec ^{2} \theta=1+\tan ^{2} \theta$
Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
$(\sin A+\operatorname{cosec} A)^{2}+(\cos A+\sec A)^{2}=7+\tan ^{2} A+\cot ^{2} A$
Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
$(\sin A+\operatorname{cosec} A)^{2}+(\cos A+\sec A)^{2}=7+\tan ^{2} A+\cot ^{2} A$